Can someone answer number 2? Thank you

C. (77.29, 85.71)

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean test score of randomly selected students.

Number of sample, n = 25

Mean, u = 81.5

Standard deviation, s = 10.2

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:

81.5 +/- 1.96 × 10.2/√25

= 81.5 +/- 1.96 × 2.04

= 81.5 +/- 3.9984

The lower end of the confidence interval is 81.5 - 3.9984 =77.5016

The upper end of the confidence interval is 81.5 + 3.9984 =85.4984

Therefore, the correct option is

C. (77.29, 85.71)

search the worksheet up on the internet

step-by-step explanation:

it works trust me

6hryt%4578855