SOLUTION: Assume that females have pulse rates that are normally distributed with a mean of = 76.0 beats per minute and a standard deviation of o = 12.5 beats per minute. Complete parts (a)

Question 1201609: Assume that females have pulse rates that are normally distributed with a mean of = 76.0 beats per minute and a standard deviation of o = 12.5 beats per minute. Complete parts (a) through (C)
k below.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 72 beats per minute and 80 beats per minute
The probability is
[Round to tour decimal o aces as needed.
b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean between 72 beats per minute and 80 beats 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 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 an sample size
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Answers:
(a) 0.2510 (approximate)
(b) 0.8904 (approximate)
(c) A. 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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Explanation:

Part (a)

mu = 76 = population mean pulse rate
sigma = 12.5 = population standard deviation of the pulse rates
x = pulse rate of 1 individual

Each pulse rate is measured in beats per minute (bpm).

Compute the z score for the raw score x = 72
z = (x-mu)/sigma
z = (72-76)/12.5
z = -4/12.5
z = -0.32

Repeat for x = 80
z = (x-mu)/sigma
z = (80-76)/12.5
z = 4/12.5
z = 0.32

The task of finding P(72 < x < 80) is equivalent to P(-0.32 < z < 0.32)

I'll be using this Z table
https://www.ztable.net/
a similar table can be found in the back of your stats textbook

Use such a table to find that
P(Z < -0.32) = 0.37448
and
P(Z < 0.32) = 0.62552

So,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-0.32 < Z < 0.32) = P(Z < 0.32) - P(Z < -0.32)
P(-0.32 < Z < 0.32) = 0.62552 - 0.37448
P(-0.32 < Z < 0.32) = 0.25104
P(-0.32 < Z < 0.32) = 0.2510

Which leads back to
P(72 < x < 80) = 0.2510

There's about a 25.1% chance of randomly selecting a woman who has a pulse rate between 72 bpm and 80 bpm.

You could use a specialized stats calculator such as this one
https://onlinestatbook.com/2/calculators/normal_dist.html
as an alternative to using a Z table.

This article goes over a few examples of how to calculate normal distribution probabilities on a TI84 calculator.
https://www.statology.org/normal-probabilities-ti-84-calculator/

-------------------------

Part (b)

n = 25 = sample size
xbar = sample mean = (add up the values)/(number of values)

Part (a) refers to selecting 1 woman, while part (b) handles selecting a random sample of 25 women.
Then we compute xbar and want to determine P(72 < xbar < 80)
We'll use the aptly named xbar distribution.

The xbar distribution has its center at mu = 76 and standard deviation of SE = 2.5, where SE stands for "standard error"
SE = standard error
SE = sigma/sqrt(n)
SE = (12.5)/(sqrt(25))
SE = 2.5

Compute the z scores.
Use xbar in place of x.
Use the SE value in place of sigma.

If xbar = 72, then,
z = (xbar-mu)/SE
z = (72-76)/(2.5)
z = -4/(2.5)
z = -1.6

If xbar = 80, then,
z = (xbar-mu)/SE
z = (80-76)/(2.5)
z = 4/(2.5)
z = 1.6

The task of finding P(72 < xbar < 80) is equivalent to P(-1.6 < z < 1.6)

Use a Z table to determine the following
P(Z < -1.6) = 0.0548
P(Z < 1.6) = 0.9452

So,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-1.6 < Z < 1.6) = P(Z < 1.6) - P(Z < -1.6)
P(-1.6 < Z < 1.6) = 0.9452 - 0.0548
P(-1.6 < Z < 1.6) = 0.8904

Which leads back to
P(72 < xbar < 80) = 0.8904

There's about an 89.04% chance of randomly selecting a group of 25 women who have a sample mean (xbar) pulse rate between 72 bpm and 80 bpm.

--------------------------

Part (c)

At the top of the problem it states "Assume that females have pulse rates that are normally distributed"
So because the population is normally distributed, it indicates the distribution of sample means (xbar distribution) is also normally distributed. This is regardless of the sample size.

Therefore, we arrive at the final answer A. 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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