Normal Distribution? continued Assume that the chest measurements in Exercise 6.90 a...

Normal Distribution? continued Assume that the chest measurements in Exercise 6.90 are normally distributed with a mean of µ = 39.83 and standard deviation of σ = 2.05.

a. What proportion of the observations would lie between 36.5 and 43.5 inches?

b. Between what two measurements would 95% of the observations lie?

c. What are the actual proportions for parts a and b using the data directly? Comment on the accuracy of the proportions found using assumed normality of the chest measurements.

Reference:

Normal Distribution? The chest measurements for 5738 Scottish militiamen in the early 19^th century are given below.9 Chest sizes are measured in inches, and each observation reports the number of soldiers with that chest size.

Notice the approximate normality of the histogram of the 5738 chest measurements.

a. The mean of this distribution is ̅x = 39.83 and the standard deviation is s = 2.05. What is the 95^th percentile of this distribution based on a normal curve with µ = 39.83 and σ = 2.05?

b. Find the empirical estimate of the 95th percentile and compare with your answer in part a. (HINT: The 95th percentile will be in position .95(n + 1) = .95 × 5739 = 5452.05 from the left tail of the distribution or in position 5738 - 5452.05 = 285.95from the right tail of the distribution.)

c. Find the 90th percentile of this distribution based on a normal curve with µ = 39.83 and σ = 2.05. What is the value of the empirical 90th percentile? How does it compare with the value assuming normality?