<p> Class worksheet #2</p><p>Sampling Distribution of the… a. Mean b. Proportion c. Difference between two variables</p><p>1. The diameters of Ping-Pong balls manufactured at a large factory are normally distributed with a mean of 1.3 inches and a standard deviation of .04 inch. a. What is the probability that a randomly selected ball will have a diameter between 1.28 and 1.32 inches? b. If many random samples of 16 Ping-Pong balls are selected, what is the probability that the mean diameter will fall in the interval 1.28 through 1.32?</p><p>2. Trucks were weighed on two scales. The WIM (weight-in-motion) scale weighs trucks as they drive down the highway. The POE scale weighs the trucks while they stop at the port-of-entry station. The makers of the WIM scale believe that their scale will weigh heavier than the POE scale 60% of the time when gross weight is considered. Create a variable (called Index) that has a value = 1 when the WIM gross weight > POE gross weight, and a value = 0 otherwise. From the data presented in the file TRUCKS… a. Determine the sample proportion of times the WIM gross weight exceeds the POE gross weight. b. Assuming the WIM makers’ assertion is correct, what is the probability of finding a proportion as extreme as or less extreme than that found in part ‘a’? c. Based on the probability found in part ‘b’, what should the WIM makers conclude? Is there 60% figure reasonable? </p><p>3. For the data provided in problem 2, what is the probability that the mean weight of the WIM weight reading will exceed the mean weight of the POE weight reading by </p><p> not more than 1500 lbs? Assume the WIM population has the parameters 1=63,330 </p><p> and 1 = 18,200, and the POE population has the parameters 2=60,150 and </p><p>2=19,750. Solutions</p><p>1. Let X represents the ball diameter. Then, X is normally distributed with  = 1.3 and  = .04. a. The probability statement: P(1.28<X<1.32) = P((1.28-1.3)/.04<Z<(1.32-1.3)/.04) =P(-.5<Z<.5) = 2P(0<Z<.5) b. The probability statement here pertains to the sample mean X . 1.28 1.3 1.32 1.3     P(1.28  X  1.32) P  Z    P(2  Z  2)  2P(0  Z  2) 2.2  .04 16 .04 16 </p><p>2. This problem deals with the sample proportion. a. Define a success as the event when Index = 1 (that is, when the WIM gross weight > POE gross weight). The required proportion can be calculated from the count of times for which Index = 1, divided by the total count of 200. The sample proportion can be easily found in Excel. Label column C (next to the WIM column) as Index. Type the following command in cell C2: =if (b2>a2, 1, 0), and drag down to cover the entire column. The numbers 1 and 0 are placed in the proper cells. If you are unsuccessful in working with this formula just enter the numbers 1 and 0 manually, based on the relationship between WIM gross weight and POE gross weight. Now use Data Analysis Plus. You can apply the option “Z-Estimate: Proportion”, and indicate (when asked) that the code for a success is ‘1’. The sample proportion of 0.53 appears as part of the output. Assuming the manufacturer assertion is correct b. P( pˆ<.53)=P(Z < (.53-.60)/[.60(1-.60)/(200)]1/2 = .02</p><p>3. The question is about the probability of the difference between the two sample means. Specifically,      1,500 (63,300 60,150) P(X1  X2  1500)  P Z   18,2002 19,7502      200 200 </p>