Question

Use the accompanying data set on the pulse rates​ (in beats per​ minute) of males to complete parts​ (a) and​ (b) below.

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Click the icon to view the pulse rates of males.

a. Find the mean and standard​ deviation, and verify that the pulse rates have a distribution that is roughly normal.

The mean of the pulse rates is

71.7

beats per minute.

​(Round to one decimal place as​ needed.)

The standard deviation of the pulse rates is

12.2

beats per minute.

​(Round to one decimal place as​ needed.)

Explain why the pulse rates have a distribution that is roughly normal. Choose the correct answer below.A.

The pulse rates have a distribution that is normal because a histogram of the data set is​ bell-shaped and symmetric

B.

The pulse rates have a distribution that is normal because the mean of the data set is equal to the median of the data set.

C.

The pulse rates have a distribution that is normal because none of the data points are greater than 2 standard deviations from the mean

D.

The pulse rates have a distribution that is normal because none of the data points are negative.

b. Treating the unrounded values of the mean and standard deviation as​ parameters, and assuming that male pulse rates are normally​ distributed, find the pulse rate separating the lowest​ 2.5% and the pulse rate separating the highest​ 2.5%. These values could be helpful when physicians try to determine whether pulse rates are significantly low or significantly high.

Answer 1

Answer

(A) For roughly normal distribution, we always need symmetric and bell shaped or mound shaped distribution

so, option A is correct

The pulse rates have a distribution that is normal because a histogram of the data set is​ bell-shaped and symmetric

(B) We will use invNorm function in TI 84 calculator

for lowest 2.5, we have Area = 0.025 with mean = 71.7 and standard deviation = 12.2

so, pulse rate separating the lowest​ 2.5% = invNorm(0.025,71.7,12.2)

= 47.8

and

area = 1 - 0.025 = 0.975 (for highest 2.5%)

with

mean = 71.7 and standard deviation = 12.2

so, pulse rate separating the highest 2.5% = invNorm(0.975,71.7,12.2)

= 95.6