A random sample of 28 male runners showed that they spent an average of 22 hours a week preparing for races. The standard deviation of the sample was 2 hours. Find the 95% confidence interval.

Step-by-step explanation:

We want to determine a 95% confidence interval for the average time spent by male runners in a week in preparing for races.

Number of sample, n = 28

Mean, u = 22 hours

Standard deviation, s = 2 hours

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

22 ± 1.96 × 2/√28

= 22 ± 1.96 × 0.378

= 22 ± 0.741

The lower end of the confidence interval is 22 - 0.741 =21.259

The upper end of the confidence interval is 22 + 0.741 =22.741

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice