Question

Assume that females have pulse rates that are normally distributed with a mean of mu equals 73.0 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 67 beats per minute and 79 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 67 beats per minute and 79 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​ individuals, not sample​ means, the distribution is a normal distribution for any sample size. B. Since the distribution is of sample​ means, not​ individuals, the distribution 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 mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.

Answer 1

Solution :

Given that ,

mean = = 73.0

standard deviation = = 12.5

a) P( 67 < x < 79) = P[(67 - 73.0)/ 12.5 ) < (x - ) /  < (79 - 73.0) / 12.5 ) ]

= P( -0.48 < z < 0.48)

= P(z < 0.48 ) - P(z < -0.48 )

Using z table,

= 0.6844 - 0.3156

= 0.3688

b) n = 4

=   = 73.0

= / n = 12.5 / 4 = 6.25

P(67 < < 79)  

= P[(67 - 73.0) / 6.25 < ( - ) / < (79 - 73.0) / 6.25)]

= P( -0.96 < Z < 0.96 )

= P(Z < 0.96) - P(Z < -0.96)

Using z table,  

= 0.8315 - 0.1685

= 0.6630

c) correct option is = C.

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