The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Lecture 8: Special Probability Densities
Assist. Prof. Dr. Emel YAVUZ DUMAN
Int. to Prob. Theo. and Stat & Int. to Probability Istanbul˙ K¨ult¨ur University Faculty of Engineering The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Outline
1 The Uniform Distribution
2 The Normal Distribution
3 The Normal Approximation to the Binomial Distribution
4 The Normal Approximation to the Poisson Distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Outline
1 The Uniform Distribution
2 The Normal Distribution
3 The Normal Approximation to the Binomial Distribution
4 The Normal Approximation to the Poisson Distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Definition 1 A random variable has a uniform distribution and it is refereed to as a continuous uniform random variable if and only if its probability density is given by
1 β α for α< x < β, u(x; α,β)= − (0 elsewhere. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Definition 1 A random variable has a uniform distribution and it is refereed to as a continuous uniform random variable if and only if its probability density is given by
1 β α for α< x < β, u(x; α,β)= − (0 elsewhere.
The parameters α and β of this probability density are real constants, with α<β, The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Definition 1 A random variable has a uniform distribution and it is refereed to as a continuous uniform random variable if and only if its probability density is given by
1 β α for α< x < β, u(x; α,β)= − (0 elsewhere.
The parameters α and β of this probability density are real constants, with α<β, and may be pictured as in the figure.
u(x; α,β)
1 β α −
x α β The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Theorem 2 Them mean and the variance of the uniform distribution are given by α + β 1 µ = and σ2 = (β α)2 2 12 − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Theorem 2 Them mean and the variance of the uniform distribution are given by α + β 1 µ = and σ2 = (β α)2 2 12 − Proof. The mean: β x µ = E(X )= ∞ xu(x; α,β)dx = dx α β α Z−∞ Z − β x2 β2 α2 (β α)(β + α) = = − = − 2(β α) 2(β α) 2(β α) − α − − β + α = 2 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The variance: β x2 E(X 2)= ∞ x2u(x; α,β)dx = dx α β α Z−∞ Z − β x3 β3 α3 (β α)(β2 + βα + α2) = = − = − 3(β α) 3(β α) 3(β α) − α − − 2 2 β + βα + α = 3 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The variance: β x2 E(X 2)= ∞ x2u(x; α,β)dx = dx α β α Z−∞ Z − β x3 β3 α3 (β α)(β2 + βα + α2) = = − = − 3(β α) 3(β α) 3(β α) − α − − 2 2 β + βα + α = 3
β2 + βα + α2 (β + α)2 σ2 = E(X 2) (E(X ))2 = − 3 − 22 4β2 + 4βα + 4α2 3β2 6αβ 3α2 = − − − 12 β2 2βα + α2 (β α)2 = − = − . 12 12 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 3 Southwest Arizona State University provides bus service to students while they are on campus. A bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 am and 11 pm during weekdays. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed from 0 to 30 minutes. (a) Draw a graph of this distribution. (b) Show that the area of this uniform distribution is 1.00. (c) How long will a student typically have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times? (d) What is the probability a student will wait more than 25 minutes? (e) What is the probability a student will wait between 10 and 20 minutes? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Solution. (a) The graph of this distribution:
Probability 1 ¯ 30 0 = 0.03 −
0 10 20 30 Length of wait (minutes) The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(b) The times students must wait for the bus is uniform over the interval from 0 minutes to 30 minutes, so in this case α = 0 and β = 30. 1 Area = (height)(base) (30 0) = 0 (30 0) − − or 30 1 Area = dx 30 0 Z0 − x 30 30 0 = = − = 1. 30 0 30
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(c) The mean waiting time: α + β 0 + 30 µ = = = 15 2 2 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(c) The mean waiting time: α + β 0 + 30 µ = = = 15 2 2 The standard deviation of the waiting time:
(β α)2 (30 0)2 σ = − = − = 5√3 = 8.6603 r 12 r 12 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(c) The mean waiting time: α + β 0 + 30 µ = = = 15 2 2 The standard deviation of the waiting time:
(β α)2 (30 0)2 σ = − = − = 5√3 = 8.6603 r 12 r 12
P(X )
0.03¯
0 10 20 30 µ = 15 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
30 1 x 30 5 1 ¯ (d) P(25 < X < 30) = 25 30 0 dx = 30 25 = 30 = 6 = 0.16 − R
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
30 1 x 30 5 1 ¯ (d) P(25 < X < 30) = 25 30 0 dx = 30 25 = 30 = 6 = 0.16 − R P(X ) Area= 0.16¯ 0.03¯
0 10 2025 30 µ = 15 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
20 1 x 20 10 1 ¯ (e) P(10 X 20) = 10 30 0 dx = 30 10 = 30 = 3 = 0.3 ≤ ≤ − R
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
20 1 x 20 10 1 ¯ (e) P(10 X 20) = 10 30 0 dx = 30 10 = 30 = 3 = 0.3 ≤ ≤ − R P(X ) Area= 0.3¯ 0.03¯
0 10 20 30 µ = 15 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Outline
1 The Uniform Distribution
2 The Normal Distribution
3 The Normal Approximation to the Binomial Distribution
4 The Normal Approximation to the Poisson Distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution, which we shall study in this section, is in many ways the cornerstone of modern statistical theory. Definition 4 A random variable X has a normal distribution and it is referred to as a normal random variable if and only if its probability density is given by
1 1 ( x−µ )2 n(x; µ,σ)= e− 2 σ for < x < σ√2π − ∞ ∞ where σ > 0. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the x-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the x-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The location of a normal distribution is determined by the mean, µ, the dispersion or spread of the distribution is determined by the standard deviation, σ. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the x-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The location of a normal distribution is determined by the mean, µ, the dispersion or spread of the distribution is determined by the standard deviation, σ. The arithmetic mean, median, and mode are equal. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the x-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The location of a normal distribution is determined by the mean, µ, the dispersion or spread of the distribution is determined by the standard deviation, σ. The arithmetic mean, median, and mode are equal. The total area under the curve is 1.00; half the area under the normal curve is to the right of this center point and the other half to the left of it. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the x-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The location of a normal distribution is determined by the mean, µ, the dispersion or spread of the distribution is determined by the standard deviation, σ. The arithmetic mean, median, and mode are equal. The total area under the curve is 1.00; half the area under the normal curve is to the right of this center point and the other half to the left of it. ∞ ∞ 1 1 ( x−µ )2 n(x; µ,σ)dx = e− 2 σ dx = 1. σ√2π Z−∞ Z−∞ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Normal Distribution - Graphically The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Equal Means and Different Means and Different Standard Deviations Standard Deviations
Different Means and Equal Standard Deviations The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Theorem 5 The moment-generating function of the normal distribution is given by µt+ 1 σ2t2 MX (t)= e 2 . The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Since the normal distribution plays a basic role in statistics and its density cannot be integrated directly, its areas have been tabulated for the special case, where µ = 0 and σ = 1. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Since the normal distribution plays a basic role in statistics and its density cannot be integrated directly, its areas have been tabulated for the special case, where µ = 0 and σ = 1. Definition 6
Let X be a random variable with mean µ∗ and standard deviation σ∗ (σ∗ > 0). Then we can define an associated standard random variable given by X µ Z = − ∗ . σ∗ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Since the normal distribution plays a basic role in statistics and its density cannot be integrated directly, its areas have been tabulated for the special case, where µ = 0 and σ = 1. Definition 6
Let X be a random variable with mean µ∗ and standard deviation σ∗ (σ∗ > 0). Then we can define an associated standard random variable given by X µ Z = − ∗ . σ∗
X µ∗ 1 1 µZ = E(Z)= E − = E(X µ∗)= [E(X ) µ∗] = 0, σ∗ σ∗ − σ∗ − µ∗
2 X µ∗ σZ = Var(Z)= E − | {z } σ ∗ 1 2 1 2 = 2 E[(X µ∗) ]= 2 E[(X µ∗) ] = 1 σZ = 1. σ∗ − σ∗ − ⇒ σ∗2 | {z } The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Definition 7 The normal distribution with µ = 0 and σ = 1 is referred to as the standard normal distribution. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Definition 7 The normal distribution with µ = 0 and σ = 1 is referred to as the standard normal distribution.
µ
The red curve is the standard normal distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The entries in the table, represented by the shaded area of the figure, are the values of z 1 − 1 x2 e 2 dx √2π Z0 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The entries in the table, represented by the shaded area of the figure, are the values of z 1 − 1 x2 e 2 dx √2π Z0 that is the probabilities that a random variable having the standard normal distribution will take on a value on the interval from 0 to z, for z = 0.00, 0.01, 0.02, , 3.08 and 3.09 and also z = 4.0, z = 5.0, and ··· z = 6.0.
1.5 1 1 x2 e 2 dx = 0.4332 0 √2π − R The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; (d) between 0.25 and 0.45. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; (d) between 0.25 and 0.45. − Solution. (a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; (d) between 0.25 and 0.45. − Solution. (a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573; (b) P(Z < 0.88) = 0.5 0.3106 = 0.1894; − − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; (d) between 0.25 and 0.45. − Solution. (a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573; (b) P(Z < 0.88) = 0.5 0.3106 = 0.1894; − − (c) P(1.3 < Z < 1.75) = 0.4599 0.4032 = 0.0567; − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 8 Find the probabilities that a random variable having the standard normal distribution will take on a value (a) less that 1.72; (b) less than 0.88; − (c) between 1.3 and 1.75; (d) between 0.25 and 0.45. − Solution. (a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573; (b) P(Z < 0.88) = 0.5 0.3106 = 0.1894; − − (c) P(1.3 < Z < 1.75) = 0.4599 0.4032 = 0.0567; − (d) P( 0.25 < Z < 0.45) = 0.0987 + 0.1736 = 0.2727. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 9 With the reference to Standard Normal Distribution Table, find the values of z that correspond to entries of (a) 0.3512; (b) 0.2533. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 9 With the reference to Standard Normal Distribution Table, find the values of z that correspond to entries of (a) 0.3512; (b) 0.2533.
Solution. (a) Since 0.3512 falls between 0.3508 and 0.3531, corresponding to z = 1.04 and z = 1.05, and since 0.3512 is closer to 0.3508 than 0.3531, we choose z = 1.04. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 9 With the reference to Standard Normal Distribution Table, find the values of z that correspond to entries of (a) 0.3512; (b) 0.2533.
Solution. (a) Since 0.3512 falls between 0.3508 and 0.3531, corresponding to z = 1.04 and z = 1.05, and since 0.3512 is closer to 0.3508 than 0.3531, we choose z = 1.04. (b) Since 0.2533 falls midway between 0.2517 and 0.2549, corresponding to z = 0.68 and z = 0.69, we choose z = 0.685. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; (d) between z and z is 0.8502. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; (d) between z and z is 0.8502. − Solution. (a) z = 1.92 ± The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; (d) between z and z is 0.8502. − Solution. (a) z = 1.92 ± (b) 0.9869 0.5 = 0.4869 z = 2.22; − ⇒ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; (d) between z and z is 0.8502. − Solution. (a) z = 1.92 ± (b) 0.9869 0.5 = 0.4869 z = 2.22; − ⇒ (c) 0.5 0.1314 = 0.3686 z = 1.12; − ⇒ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 10 Find z if the standard-normal-curve area (a) between 0 and z is 0.4726; (b) to the left of z is 0.9869; (c) to the right of z is 0.1314; (d) between z and z is 0.8502. − Solution. (a) z = 1.92 ± (b) 0.9869 0.5 = 0.4869 z = 2.22; − ⇒ (c) 0.5 0.1314 = 0.3686 z = 1.12; − ⇒ (d) 0.8502 2 = 0.4251 z = 1.44. ÷ ⇒ ± The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
To determine probabilities related to random variables having normal distribution other than the standard normal distribution, we make use of the following theorem: The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
To determine probabilities related to random variables having normal distribution other than the standard normal distribution, we make use of the following theorem: Theorem 11 If X has a normal distribution with mean µ and the standard deviation σ, then X µ Z = − σ has the standard normal distribution. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
To determine probabilities related to random variables having normal distribution other than the standard normal distribution, we make use of the following theorem: Theorem 11 If X has a normal distribution with mean µ and the standard deviation σ, then X µ Z = − σ has the standard normal distribution. Thus, to use the Standard Normal Distribution Table in connection with any random variable having a normal distribution, we simply x µ perform the change of scale z = −σ . The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 12 Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across the US is a random variable having a normal distribution with mean of 4.35 mrem and a standard deviation of 0.59 mrem. What is the probability that a person will be exposed to more that 5.20 mrem of cosmic radiation on such a flight? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 12 Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across the US is a random variable having a normal distribution with mean of 4.35 mrem and a standard deviation of 0.59 mrem. What is the probability that a person will be exposed to more that 5.20 mrem of cosmic radiation on such a flight?
Solution. Looking up the entry corresponding to 5.20 4.35 z = − = 1.44 0.59 in the Standard Normal Distribution Table and subtracting it from 0.5, we get 0.5 0.4251 = 0.0749. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 13 Suppose that during periods of transcendental meditation the reduction of a person’s oxygen consumption is a random variable having a normal distribution with µ = 37.6 cc per minute and σ = 4.6 cc per minute. Find the probabilities that during a period of transcendental meditation a person’s oxygen consumption will be reduced by The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 13 Suppose that during periods of transcendental meditation the reduction of a person’s oxygen consumption is a random variable having a normal distribution with µ = 37.6 cc per minute and σ = 4.6 cc per minute. Find the probabilities that during a period of transcendental meditation a person’s oxygen consumption will be reduced by (a) at least 44.5 cc per minute; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 13 Suppose that during periods of transcendental meditation the reduction of a person’s oxygen consumption is a random variable having a normal distribution with µ = 37.6 cc per minute and σ = 4.6 cc per minute. Find the probabilities that during a period of transcendental meditation a person’s oxygen consumption will be reduced by (a) at least 44.5 cc per minute; (b) at most 35.0 cc per minute; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 13 Suppose that during periods of transcendental meditation the reduction of a person’s oxygen consumption is a random variable having a normal distribution with µ = 37.6 cc per minute and σ = 4.6 cc per minute. Find the probabilities that during a period of transcendental meditation a person’s oxygen consumption will be reduced by (a) at least 44.5 cc per minute; (b) at most 35.0 cc per minute; (c) anywhere from 30.0 to 40.0 cc per minute. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Solution. (a) at least 44.5 cc per minute: Looking up the entry corresponding to 44.5 37.6 Z = − = 1.5 4.6 in the Standard Normal Distribution Table and subtracting it from 0.5, we get P(Z 1.5) = 0.5 0.4332 = 0.0668. ≥ − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Solution. (a) at least 44.5 cc per minute: Looking up the entry corresponding to 44.5 37.6 Z = − = 1.5 4.6 in the Standard Normal Distribution Table and subtracting it from 0.5, we get P(Z 1.5) = 0.5 0.4332 = 0.0668. ≥ −
(b) at most 35.0 cc per minute: Looking up the entry corresponding to 35.0 37.6 Z = − = 0.565 4.6 − in the Standard Normal Distribution Table as 0.2123+0.2157 2 = 0.2140 and subtracting it from 0.5, we get P(Z 0.565) = 0.5 0.2140 = 0.2860. ≤− − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(c) anywhere from 30.0 to 40.0 cc per minute: Looking up the entries corresponding to 30 37.6 Z1 = − = 1.6522 1.65 4.6 − ≈− and 40 37.6 Z2 = − = 0.5217 0.52 4.6 ≈ in the Standard Normal Distribution Table, we obtain that
P( 1.65 Z 0.52) = 0.4505 + 0.1985 = 0.6490. − ≤ ≤ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 14 A random variable has a normal distribution with σ = 10. If the probability that the random variable will take on a value less than 82.5 is 0.8212, what is the probability that it will take on a value greater than 58.3? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 14 A random variable has a normal distribution with σ = 10. If the probability that the random variable will take on a value less than 82.5 is 0.8212, what is the probability that it will take on a value greater than 58.3?
Solution.
Total area: 0.8212 Area of rhs: 0.3212
82.5 µ 0.92 = 10−
Since 0.8212 > 0.5, we see that z lies on the right hand side of the mean. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 14 A random variable has a normal distribution with σ = 10. If the probability that the random variable will take on a value less than 82.5 is 0.8212, what is the probability that it will take on a value greater than 58.3?
Solution.
Total area: 0.8212 Area of rhs: 0.3212
82.5 µ 0.92 = 10−
Since 0.8212 > 0.5, we see that z lies on the right hand side of the mean. Then, 0.8212 0.5 = 0.3212 is the area between 0 and z. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 14 A random variable has a normal distribution with σ = 10. If the probability that the random variable will take on a value less than 82.5 is 0.8212, what is the probability that it will take on a value greater than 58.3?
Solution.
Total area: 0.8212 Area of rhs: 0.3212
82.5 µ 0.92 = 10−
Since 0.8212 > 0.5, we see that z lies on the right hand side of the mean. Then, 0.8212 0.5 = 0.3212 is the area between 0 and z. − By using the Standard Normal Distribution Table we obtain that z = 0.92. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Therefore 82.5 µ 0.92 = − µ = 73.3. 10 ⇒ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Therefore 82.5 µ 0.92 = − µ = 73.3. 10 ⇒ Thus the probability that it will take on a value greater than 58.3 is 58.3 73.3 z = − = 1.5 10 − ⇒ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Therefore 82.5 µ 0.92 = − µ = 73.3. 10 ⇒ Thus the probability that it will take on a value greater than 58.3 is 58.3 73.3 z = − = 1.5 10 − ⇒
P(Z > 1.5) = 0.5 + 0.4332 = 0.9332. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Outline
1 The Uniform Distribution
2 The Normal Distribution
3 The Normal Approximation to the Binomial Distribution
4 The Normal Approximation to the Poisson Distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
We have already seen that the Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of θ provided that the correct conditions exist. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
We have already seen that the Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of θ provided that the correct conditions exist. The approximation is only of practical use if just a few terms of the Poisson distribution need be calculated. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
We have already seen that the Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of θ provided that the correct conditions exist. The approximation is only of practical use if just a few terms of the Poisson distribution need be calculated. In cases where many - sometimes several hundred - terms need to be calculated the arithmetic involved becomes very tedious indeed and we turn to the normal distribution for help. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution is sometimes introduced as a continuous distribution that provides a close approximation to the binomial distribution when n, the number of trials, is very large and θ, the probability of success on an individual trials, is close to 1/2. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution is sometimes introduced as a continuous distribution that provides a close approximation to the binomial distribution when n, the number of trials, is very large and θ, the probability of success on an individual trials, is close to 1/2. The graph of the binomial distribution where n = 50 and θ = 0.5 is shown in the following figure.
Notice that it approximates a normal distribution. This suggests that a binomial distribution can be approximated by a normal distribution as long as the number of trials is relatively large. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Theorem 15 If X is a random variable having a binomial distribution with the parameters n and θ, then the moment-generating function of X nθ Z = − nθ(1 θ) − approaches that of the standardp normal distribution when n . → ∞ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Theorem 15 If X is a random variable having a binomial distribution with the parameters n and θ, then the moment-generating function of X nθ Z = − nθ(1 θ) − approaches that of the standardp normal distribution when n . → ∞ Remark. If X is a binomial random variable of n independent trials, each with probability of success θ, and if
nθ> 5 and n(1 θ) > 5 − then the binomial random variable can be approximated by a normal distribution with
µ = nθ and σ = nθ(1 θ). − p The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe the number of heads. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe the number of heads. Suppose we want to compute the probability of obtaining exactly 4 heads. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe the number of heads. Suppose we want to compute the probability of obtaining exactly 4 heads. Whereas a discrete random variable can have only a specified value (such as 4), a continuous random variable used to approximate it could take on any values within an interval around that specified value, The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe the number of heads. Suppose we want to compute the probability of obtaining exactly 4 heads. Whereas a discrete random variable can have only a specified value (such as 4), a continuous random variable used to approximate it could take on any values within an interval around that specified value, as demonstrated in this figure: The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe the number of heads. Suppose we want to compute the probability of obtaining exactly 4 heads. Whereas a discrete random variable can have only a specified value (such as 4), a continuous random variable used to approximate it could take on any values within an interval around that specified value, as demonstrated in this figure: The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The continuity correction requires adding or subtracting 0.5 from the value or values of the discrete random variable X as needed. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The continuity correction requires adding or subtracting 0.5 from the value or values of the discrete random variable X as needed. Hence to use the normal distribution to approximate the probability of obtaining exactly 4 heads (i.e., X = 4), we would find the area under the normal curve from X = 3.5 to X = 4.5, the lower and upper boundaries of 4. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Moreover, to determine the approximate probability of observing at least 4 heads, we would find the area under the normal curve from X = 3.5 and above since, on a continuum, 3.5 is the lower boundary of X . The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Moreover, to determine the approximate probability of observing at least 4 heads, we would find the area under the normal curve from X = 3.5 and above since, on a continuum, 3.5 is the lower boundary of X . Similarly, to determine the approximate probability of observing at most 4 heads, we would find the area under the normal curve from X = 4.5 and below since, on a continuum, 4.5 is the upper boundary of X . The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· We also know that the mean of this binomial distribution is given by µ = nθ = 12(0.5) = 6, The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· We also know that the mean of this binomial distribution is given by µ = nθ = 12(0.5) = 6, We also know that the standard deviation of this binomial distribution is given by σ = nθ(1 θ)= 12(0.5)(0.5) = 1.732. − p p The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· We also know that the mean of this binomial distribution is given by µ = nθ = 12(0.5) = 6, We also know that the standard deviation of this binomial distribution is given by σ = nθ(1 θ)= 12(0.5)(0.5) = 1.732. − Say we are interestedp in the probabilityp of observing between 3 and 5 heads, inclusive; The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· We also know that the mean of this binomial distribution is given by µ = nθ = 12(0.5) = 6, We also know that the standard deviation of this binomial distribution is given by σ = nθ(1 θ)= 12(0.5)(0.5) = 1.732. − Say we are interestedp in the probabilityp of observing between 3 and 5 heads, inclusive; that is, P(3 X 5). ≤ ≤ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Now consider the binomial distribution in particular. Let X be the number of heads in 12 flips of a fair coin. We know that the probability of observing exactly x heads in 12 flips is
12 x 12 x P(X = x)= b(x; 12, 0.5) = 0.5 0.5 − , x = 0, 1, 2 , 12. x ··· We also know that the mean of this binomial distribution is given by µ = nθ = 12(0.5) = 6, We also know that the standard deviation of this binomial distribution is given by σ = nθ(1 θ)= 12(0.5)(0.5) = 1.732. − Say we are interestedp in the probabilityp of observing between 3 and 5 heads, inclusive; that is, P(3 X 5). We can calculate this ≤ ≤ exactly, of course: P(3 X 5) = P(X = 3)+ P(X = 4)+ P(X = 5) ≤ ≤ = 0.05371 + 0.12085 + 0.19336 = 0.36792. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. A useful rule of thumb is that the normal approximation should work well enough if both nθ and n(1 θ) are − greater than 5. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. A useful rule of thumb is that the normal approximation should work well enough if both nθ and n(1 θ) are − greater than 5. For this example, both equal 12(0.5) = 6, so we are about at the limit of usefulness of the approximation. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. A useful rule of thumb is that the normal approximation should work well enough if both nθ and n(1 θ) are − greater than 5. For this example, both equal 12(0.5) = 6, so we are about at the limit of usefulness of the approximation. Back to the question at hand. Since X is a binomial random variable, the following statement (based on the continuity correction) is exactly correct:
P(3 X 5) = P(2.5 < X < 5.5). ≤ ≤ The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. A useful rule of thumb is that the normal approximation should work well enough if both nθ and n(1 θ) are − greater than 5. For this example, both equal 12(0.5) = 6, so we are about at the limit of usefulness of the approximation. Back to the question at hand. Since X is a binomial random variable, the following statement (based on the continuity correction) is exactly correct:
P(3 X 5) = P(2.5 < X < 5.5). ≤ ≤ Note that this statement is not an approximation - it is exactly correct! The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
What about the normal approximation to this value? First, we should check whether the normal approximation is likely to work very well in this case. A useful rule of thumb is that the normal approximation should work well enough if both nθ and n(1 θ) are − greater than 5. For this example, both equal 12(0.5) = 6, so we are about at the limit of usefulness of the approximation. Back to the question at hand. Since X is a binomial random variable, the following statement (based on the continuity correction) is exactly correct:
P(3 X 5) = P(2.5 < X < 5.5). ≤ ≤ Note that this statement is not an approximation - it is exactly correct! The reason for this is that we are adding the events 2.5 < X < 3 and 5 < X < 5.5 to get from the left side of the equation to the right side of the equation, but for the binomial random variable, these events have probability zero. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The continuity correction is not where the approximation comes in; that comes when we approximate X using a normal distribution with mean µ = 6 and standard deviation σ = 1.732:
P(3 X 5) = P(2.5 < X < 5.5) ≤ ≤ 2.5 6 5.5 6 P − < Z < − ≈ 1.732 1.732 = P( 2.02 < Z < 0.29) − − = 0.4783 0.1141 = 0.3642. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The continuity correction is not where the approximation comes in; that comes when we approximate X using a normal distribution with mean µ = 6 and standard deviation σ = 1.732:
P(3 X 5) = P(2.5 < X < 5.5) ≤ ≤ 2.5 6 5.5 6 P − < Z < − ≈ 1.732 1.732 = P( 2.02 < Z < 0.29) − − = 0.4783 0.1141 = 0.3642. − Note that the approximation is only off by about 1%, which is pretty good for such a small sample size. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 16 Suppose that a sample of n = 1, 600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample not more than 150 tires will be defective? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 16 Suppose that a sample of n = 1, 600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 16 Suppose that a sample of n = 1, 600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The mean and the standard deviation are µ = nθ = 1, 600(0.08) = 128(> 5)(n(1 θ) = 1, 600(0.92) = 1, 472 > 0), − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 16 Suppose that a sample of n = 1, 600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The mean and the standard deviation are µ = nθ = 1, 600(0.08) = 128(> 5)(n(1 θ) = 1, 600(0.92) = 1, 472 > 0), −
σ = nθ(1 θ)= 1, 600(0.08)(0.92) = 10.85. − p p The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 16 Suppose that a sample of n = 1, 600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The mean and the standard deviation are µ = nθ = 1, 600(0.08) = 128(> 5)(n(1 θ) = 1, 600(0.92) = 1, 472 > 0), −
σ = nθ(1 θ)= 1, 600(0.08)(0.92) = 10.85. − The probabilityp calculation isp thus P(X 150) = P(X < 150.5) ≤ 150.5 128 P Z < − = P(Z < 2.07) ≈ 10.85 = 0.5 + 0.4808 = 0.9808. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? (c) between 20 and 25 (inclusive) are in error? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? (c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? (c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The mean and the standard deviation are
µ = nθ = 400(0.07) = 28, The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? (c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The mean and the standard deviation are
µ = nθ = 400(0.07) = 28,
σ = nθ(1 θ)= 400(0.07)(0.93) = 5.103. − p p The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 17 Based on past experience, 7% of all luncheon tickets are in error. If a random sample of 400 tickets is selected, what is the approximate probability that (a) exactly 25 are in error? (b) at most 25 are in error? (c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The mean and the standard deviation are
µ = nθ = 400(0.07) = 28,
σ = nθ(1 θ)= 400(0.07)(0.93) = 5.103. − The probabilityp calculations arep thus The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(a) exactly 25 are in error?
P(X = 25) = P(24.5 < X < 25.5) 24.5 28 25.5 28 P − < Z < − ≈ 5.103 5.103 = P( 0.69 < Z < 0.49) − − = 0.2549 0.1879 = 0.067 − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(a) exactly 25 are in error?
P(X = 25) = P(24.5 < X < 25.5) 24.5 28 25.5 28 P − < Z < − ≈ 5.103 5.103 = P( 0.69 < Z < 0.49) − − = 0.2549 0.1879 = 0.067 −
(b) at most 25 are in error?
P(X 25) = P(X < 25.5) ≤ 25.5 28 P Z < − ≈ 5.103 = P(Z < 0.49) = 0.5 0.1879 = 0.3121. − − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
(c) between 20 and 25 (inclusive) are in error?
P(20 X 25) = P(19.5 < X < 25.5) ≤ ≤ 19.5 28 25.5 28 P − < Z < − ≈ 5.103 5.103 = P( 1.67 < Z < 0.49) − − = 0.4525 0.1879 = 0.2646. − The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 18 An airline knows that over the long run, 90% of passengers who reserve seats will show up for their fights. On a particular fight with 300 seats, the airline accepts 324 reservations. Use the normal approximation to the binomial distribution to find the chance that the fight will be overbooked? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 18 An airline knows that over the long run, 90% of passengers who reserve seats will show up for their fights. On a particular fight with 300 seats, the airline accepts 324 reservations. Use the normal approximation to the binomial distribution to find the chance that the fight will be overbooked?
Solution. Let X be the number of passengers. Since µ = 324(0.9) = 291.6 and σ = 324(0.9)(0.1) = 5.4, then
P(X 301) = P(Xp> 300.5) ≥ 300.5 291.6 P Z > − ≈ 5.4 = P(Z > 1.65) = 0.5 0.4505 − = 0.0495 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 19 IQ scores of people around the world are normally distributed, with a mean of 100 and a standard deviation of 15. A genius is someone with an IQ greater than or equal to 140. What percent of the population is considered genius? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 19 IQ scores of people around the world are normally distributed, with a mean of 100 and a standard deviation of 15. A genius is someone with an IQ greater than or equal to 140. What percent of the population is considered genius?
Solution.
P(X 140) = P(X > 139.5) ≥ 139.5 100 P Z > − ≈ 15 = P(Z > 2.63) = 0.5 0.4957 − = 0.0043 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation Outline
1 The Uniform Distribution
2 The Normal Distribution
3 The Normal Approximation to the Binomial Distribution
4 The Normal Approximation to the Poisson Distribution The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Poisson distribution is also a discrete random variable, whereas the normal distribution is continuous. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Poisson distribution is also a discrete random variable, whereas the normal distribution is continuous. Therefore, we need to take this into account when we are using the normal distribution to approximate a Poisson using a continuity correction. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Poisson distribution is also a discrete random variable, whereas the normal distribution is continuous. Therefore, we need to take this into account when we are using the normal distribution to approximate a Poisson using a continuity correction. If X is a Poisson random variable with µ = λ and σ2 = λ, X λ Z = − √λ is approximately a standard normal variable. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Poisson distribution is also a discrete random variable, whereas the normal distribution is continuous. Therefore, we need to take this into account when we are using the normal distribution to approximate a Poisson using a continuity correction. If X is a Poisson random variable with µ = λ and σ2 = λ, X λ Z = − √λ is approximately a standard normal variable. The same continuity correction used for the binomial distribution can also be applied. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
The normal distribution can also be used to approximate the Poisson distribution for large values of λ. The Poisson distribution is also a discrete random variable, whereas the normal distribution is continuous. Therefore, we need to take this into account when we are using the normal distribution to approximate a Poisson using a continuity correction. If X is a Poisson random variable with µ = λ and σ2 = λ, X λ Z = − √λ is approximately a standard normal variable. The same continuity correction used for the binomial distribution can also be applied. The approximation is good for
λ> 5. The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 20 Assume that the number of asbestos particles in a squared meter of dust on a surface fallows a Poisson distribution with a mean of 1000. If a squared meter of dust is analyzed, what is the probability that 950 or fewer particles are found? The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 20 Assume that the number of asbestos particles in a squared meter of dust on a surface fallows a Poisson distribution with a mean of 1000. If a squared meter of dust is analyzed, what is the probability that 950 or fewer particles are found?
Solution. This probability can be expressed exactly as 950 e 10001000x P(X 950) = − . ≤ x! x=0 X The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
Example 20 Assume that the number of asbestos particles in a squared meter of dust on a surface fallows a Poisson distribution with a mean of 1000. If a squared meter of dust is analyzed, what is the probability that 950 or fewer particles are found?
Solution. This probability can be expressed exactly as 950 e 10001000x P(X 950) = − . ≤ x! x=0 X The computational difficulty is clear. The probability can be approximated as P(X 950) = P(X 950.5) ≤ ≤ 950.5 1000 P Z − ≈ ≤ √ 1000 0.4406 + 0.4418 = P(Z 1.565) = 0.5 = 0.0588. ≤− − 2 The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approximation
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