Question

Assume that females have pulse rates that are normally distributed with a mean of 72.0 beats per minute and a standard deviation of 12.5 beats per minute. Complete parts a through c below.

A. if 1 female is randomly selected, find the probability that her pulse rate is between 68 beats per minute amd 76 beats per minute.

B. If 4 adult females are selected, find the probability that they have pulse rates with a mean between 68 beats per minute and 76 beats per minute.

C. Why can the normal distribution be used in heartbeat even the sample side does not exceed 30?

Answer 1

Given: = 72, = 12.5

To find the probability, we need to find the Z scores first.

(a) Z = (X - )/ [/Sqrt(n)]. Since n = 1, Z = (X - )/

For P (68 < X < 76) = P(X < 76) – P(X < 68)

For P( X < 76)

Z = (76 – 72)/12.5 = 0.32

The probability for P(X < 76) from the normal distribution tables is = 0.6255

For P( X < 68)

Z = (68 – 72)/12.5 = -0.32

The probability for P(X < 68) from the normal distribution tables is = 0.3745

Therefore the required probability is 0.6255 – 0.3745 = 0.2510

(b) n =4, To calculate P (76 < X < 68) = P(X < 72) – P(X < 68)

For P( X < 76)

Z = (76 – 72)/[12.5/Sqrt(4)] = 0.64

The probability for P(X < 76) from the normal distribution tables is = 0.7389

For P( X < 68)

Z = (68 – 72)/[12.5/Sqrt(4)] = -0.64

The probability for P(X < a) from the normal distribution tables is = 0.2611

Therefore the required probability is 0.7389 – 0.2611 = 0.4778

(c) The normal distribution can be used since the original population has a normal distribution, and by the central limit theorem, if the original population is normally distributed, then distribution of sample mean is normal for any sample size.