Question

Assume that females have pulse rates that are normally distributed with a mean of mu equals 76.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts​ (a) through​ (c) below.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 69 beats per minute and 83 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.)

b. If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 69 beats per minute and 83 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.)

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

A. Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size.

B. Since the mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.

C. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

D. Since the distribution is of​ individuals, not sample​ means, the distribution is a normal distribution for any sample size.

Answer 1

a)

Data given:
Mean = 76, SD = 12.5, P(69 < Y < 83) = ?;

P(69 < Y < 83) = P(69 - mean < Y - mean < 83 - mean)
                  = P((69 - mean)/SD < (Y - mean)/SD < (83 - mean)/SD)
                  = P((69 - mean)/SD < Z < (83 - mean)/SD)
                  = P((69 - 76)/12.5< Z < (83 - 76)/12.5)
                  = P(-0.56 < Z < 0.56)
                  = P(Z < 0.56) - P(Z <-0.56)
                  = 0.4245

b)

Data given:
Mean = 76, SD/root(N) = 6.25, P(69 < Y < 83) = ?;

P(69 < Y < 83) = P(69 - mean < Y - mean < 83 - mean)
                  = P((69 - mean)/(SD/root(N)) < (Y - mean)/(SD/root(N)) < (83 - mean)/(SD/root(N)))
                  = P((69 - mean)/(SD/root(N)) < Z < (83 - mean)/(SD/root(N)))
                  = P((69 - 76)/6.25< Z < (83 - 76)/6.25)
                  = P(-1.12 < Z < 1.12)
                  = P(Z < 1.12) - P(Z <-1.12)
                  = 0.7373

c) Option C

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.