Construct a 95% confidence interval for the population mean, µ. assume the population has a normal distribution. a sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2.

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean mean test score of 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 score ×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 =77.5

The upper end of the confidence interval is 81.5 + 3.9984 = 85.5

Therefore, with 95% confidence interval, the mean test score of students is between 77.5 and 85.5

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s