Question

CASE STUDY: Chest Sizes of Scottish Militiamen (p. 306):

Chest Size	Frequency
33	3
34	19
35	81
36	189
37	409
38	753
39	1062
40	1082
41	935
42	646
43	313
44	168
45	50
46	18
47	3
48	1

1. The population mean and population standard deviation of the chest circumferences are 39.85 and 2.07, respectively, identify the normal curve that should be used for the chest circumferences
2. Use the table above to find the percentage of militiamen in the survey with chest circumference between 36 and 41 inches, inclusive. Note: as the circumferences were rounded to the nearest inch, you are actually finding the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches
3. Use the normal curve you identified in part (a) to obtain an approximation to the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches. Compare your answer to the exact percentage found in part (b).

Answer 1

Solution:

a)

So the normal curve we can use is Normal( 39.85, 2.07)

b)

Number of militiamen with chest circumference between 36 and 41 inches is : N

N = 189 + 409 + 753 + 1062 + 1082 + 935 = 4430

Total number of militiamen = 5732

Required percentage = N / Total number of militiamen = 4430 / 5732 = 0.7728 = 77.28%

c)

Required probability is thus

(using cumulative probability distribution we can write this as shown)

.............................(1)   

Now we will find corresponding Z score for the value of x using the identity given below

so Z score for x = 35.5    

Z score for x = 41.5

Required probability thus can now be written with respect to Z score as follows:

P(Z < -2.1014) = 0.0174

P(Z < 0.7971) = 1 - P(Z < -0.7971) = 0.7852 ( Since the normal curve is symmetrical )

Required probability = 0.7852 - 0.0174 = 0.7678  

Required percentage = 76.78%

Comparing the exact percentage 77.28% with estimate percentage 76.78%, we find that both are approximately equal, thus normal approximation in this problem is very appropriate.