Question

In: Statistics and Probability

Assume that females have pulse rates that are normally distributed with a mean of mu equals...

Assume that females have pulse rates that are normally distributed with a mean of mu equals μ=74.0 beats per minute and a standard deviation of sigma equals σ=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 less than 77 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.) b. If 44 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 77 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 original population has a normal​ distribution, the distribution of sample means 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 distribution is of​ individuals, not sample​ means, the distribution is a normal distribution for any sample size. D. Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size.

Solutions

Expert Solution

It is given that population mean and population standard deviation

(A) Probability that her pulse rate is less than 77 beats per minute

Using normalcdf(lower bound, upper bound, mean, standard deviation)

using lower bound = negative infinity or -E99, upper bound = 77, mean = 74 and standard deviation = 12.5

this implies

= normalcdf(-E99,77,74,12.5)

= 0.5948

(B) Probability that pulse rate is less than 77 beats per minute for a sample of 44

sample standard deviation =

Using normalcdf(lower bound, upper bound, mean, standard deviation)

using lower bound = negative infinity or -E99, upper bound = 77, mean = 74 and standard deviation = 1.88

this implies

= normalcdf(-E99,77,74,1.88)

= 0.9443

(C) It is clear that the population standard deviation is known and equal to 12.5, so distribution of sample means is a normal distribution for any sample size

Option A is correct

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


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