Lesson 6: Working with Grouped

Weighted (def) – an average in which some data values are assigned more importance than others.

To find the weighted mean:

1. Multiply each data value xi by its respective weight wi.

2. Sum these products.

3. Divide the result by the sum of the weights:

wiinn xw xwxwx1122 ... xw  wwwwin12...

Example

The syllabus for the Introduction to Management course at a local college specifies that the midterm exam is worth 30%, the term paper is worth 20%, and the final exam is worth 50% of your course grade. Let’s say you got a 20 on the midterm, a 70 on the term paper and a 90 on the final exam. Calculate your course average, that is the weighted sum of your grades.

 The data values are

x1=20, x2=70, and x3 =90.

 The weights are w1 =0.30, w2=0.20, and w3 =0.50.

 Weighted mean is:

Answer:

Example The following table gives the hourly wage for occupation “computer and information systems management” as reported by the U.S. Bureau of Labor for the top paying states in May of 2006. Find the average hourly wage for computer and information systems managers in these five states.

The data values are: x1=$60.32, x2=$60.25, x3=$59.39, x4=$57.98, x5=$55.95 The weights are: w1=12,380, w2=18,580, w3=9,540, w4=35,550, w5=10,130. The weighted mean is given by:

Data Set (def) – raw data obtained from the sample.

Ex: 77 89 84 83 80 80 83 82 85 92

87 88 87 86 99 93 79 83 81 78

Grouped Data Set (def) – data obtained from the sample that has already been summarized using a distribution table, a , etc.

Ex:

Example

There were 1,150 children adopted in the state of Georgia in 2006, according to the Administration for Children and Families. The of the ages of the children at adoption is given in a table below. Find the class midpoints.

Class: Age Frequency (f ) 0-0.99 12 1-5.99 611 6-10.99 320

11-15.99 161 16-17.99 46 Total =1150

Estimated Mean for Data Grouped into a Frequency Distribution

Given a frequency distribution with k classes, the estimated mean for the data set is given by m f  m f ... m f x  1 1 2 2 k k n

 n represents the total sample size, which is the sum of the class frequencies.

Example : Calculate the estimated mean age of the adopted children from example.

Let’s calculate the estimated mean:

m f  m f ... m f x  1 1 2 2 k k n

(0.5)(12)  (3.5)(611)  (8.5)(320)  (13.5)(161)  (17)(46)   6.8 1150

Answer: The estimated mean age of the children adopted in Georgia in 2006 is 6.8 years.

Estimated and for Data Grouped into a Frequency Distribution

Given a frequency distribution with k classes, sample size n, and estimated mean x

VARIANCE OF GROUPED DATA SET: (m  x)2 f  (m  x)2 f ... (m  x)2 f  2  1 1 2 2 k k n

STANDARD DEVIATION OF GROUPED DATA SET:

   2

Example : Calculate the estimated variance and the estimated standard deviation for the ages of adopted children from example last.

 Let’s calculate the estimated variance. The formula for variance is :

So the estimated variance is:

Example The following frequency distribution contains the total points scored during the 2006 regular season by the teams in the National Football League. Find the estimated mean and standard deviation of the number of points scored.

What is N? n = 32 (4+5+10+6+7=32) First, find the midpoints of each class. Points (Classes) Midpoints (m) Teams (frequencies f =) 140-259.99 200 4 260-299.99 280 5 300-339.99 320 10 340-379.99 360 6 380-499.99 440 7

Then we can calculate the estimated mean of the data set represented:

On average, the NFL teams scored about 332.5 points in the 2006 season.

Then we can calculate the estimated variance standard deviation of the data set represented:

Variance: Standard Deviation:

Points (Classes) Midpoints (m) Teams (frequencies f =) (m- x_hat) ² (m- x_hat) ² * f 140-259.99 200 4 (200- 332.5) ² = 17556.25 70225

260-299.99 280 5 (280- 332.5) ² = 2756.25 13781.25 300-339.99 320 10 (320- 332.5) ² = 156.25 1562.5 340-379.99 360 6 (360- 332.5) ² =756.25 4537.5

380-499.99 440 7 (440- 332.5) ² = 11556.25 80893.75

The number of points each NFL team scored typically differs from the mean number of points of 332.5 by about 73.10 points.