Question

Assume that females have pulse rates that are normally distributed with a mean ofμ=74.0beats per minute and a standard deviation of σ=12.5beats 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 67 beats per minute and 81 beats per minute.

The probability is ____________. ​(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 67 beats per minute and 81beats per minute.

The probability is __________ ​(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 original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

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

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

Answer 1

Given :

Assume that females have pulse rates that are normally distributed with a mean of μ=74.0 beats per minute and a standard deviation of σ=12.5 beats per minute.

Mean = = 74.0

Standard deviation = = 12.5

X ~ Normal ( = 74, =12.5)

a) If 1 adult female is randomly​ selected, the probability that her pulse rate is between 67 beats per minute and 81 beats per minute.

The probability is _

P( 67< x < 81) = P((67-74.0)/12.5 ) < (x - ) / < (81-74.0) /12.5 ) )

= P(-0.56 < z <0.56 )

= P(z < 0.56) - P(z < -0.56)

= 0.71226 - 0.2877

= 0.4246

The probability is 0.4246

b) If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 67 beats per minute and 81beats per minute.

= 74.0

= / n = 12.5 / 4 = 6.25

P(67 < <81) = P((67-74.0) /6.25) <( - ) / < (81-74.0) / 6.25 ))

= P(-1.12 < Z 1.12)

= P(Z < 1.12) - P(Z < -1.12)

= 0.8686 - 0.1314

= 0.7372

The probability is = 0.7372

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

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