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2 years ago

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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 69 beats per minute and 77 beats per minute. The probability is nothing. (Round to four decimal places as needed.)

1 answer:

slava [35]2 years ago

4 0

Answer: 0.2510

Step-by-step explanation:

Given : Mean = $\mu=\text{73.0 beats per minute}$

Standard deviation : $\sigma=\text{12.5 beats per minute}$

Sample size : n=1

We assume that females have pulse rates that are normally distributed.

Then , the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x =69

$z=\dfrac{69-73}{\dfrac{12.5}{\sqrt{1}}}=-0.32$

For x= 77

$z=\dfrac{77-73}{\dfrac{12.5}{\sqrt{1}}}=0.32$

The p-value = $P(-0.32$

$=P(z$

Hence, the probability that her pulse rate is between 69 beats per minute and 77 beats per minute =0.2510

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