Question

1 Assume that females have pulse rates that are normally distributed with a mean of p = 73.0 beats per minute and a standard deviation of 6 = 12.5 beats per minute. Complete parts (a) through (C) below a. 11 1 adult female is randomly selected, find the probability that her pulse rate is less than 79 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 less than 79 beats per minute m 1 WE os The probability is (Round to four decimal places as needed.) 2 Cal licati c. Why can the normal distribution be used in part b). even though the sample size does not exceed 30? spartios A Since the distribution is of individuals not sample means the distribution is a normal distribution for any sample size

Thank you

Answer 1

1)

Given :

u = 73,0 = 12.5

Solution :

a)

If 1 adult female is randomly selected, the probability that her pulse rate is less than 79 beats per minute will be

T= 79

$P(x < 79) = P\left ( \frac{x - \mu }{\sigma } < \frac{79-73}{12.5} \right )$

Pr < 79) = P(Z < 0.48)

Using Standard Normal Distribution Table,

Pr < 79) = 0.6844

The probability is 0.6844.

b)

If 4 adult females are randomly selected, the probability that they have pulse rate with a mean less than 79 beats per minute is

$P(\bar{x} < 79) = P\left ( \frac{\bar{x} - \mu }{\frac{\sigma}{\sqrt{n}} } < \frac{79-73}{\frac{12.5}{\sqrt{2}}} \right )$

$P(\bar{x} < 79) = P\left ( z< 0.96\right )$

Using Standard Normal Distribution Table,

The probability is 0.8315.

c)

Option B is correct.

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

2)

Given :

$\widehat{p} = \frac{279}{310} = 0.9$

$(1 - \widehat{p} ) = (1 - 0.9) = 0.1$

By using z table for CI 99%.

Solution :

99% Confidence Interval estimate for the percentage of girls born is

$\widehat{p} \pm z_{0.005}\sqrt{\frac{\widehat{p}(1 - \widehat{p})}{n}}$

$0.9 \pm 2.58\sqrt{\frac{0.9(1 - 0.9)}{310}}$

$0.9 \pm 0.045$

0.855 < p < 0.945

Does the method appear to be effective?

Yes, the proportion of the girls is significantly different from 0.5

**Thank you for giving me the opportunity to help you. Please give me thumbs up if you like this answer.**