Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students. A. 91.68 B. 92.03 < μ < 97.97
C. 92.95 D. 91.69 < μ < 98.31

B) 92.03 < μ < 97.97

99% confidence interval for the mean score of all students.

92.03 < μ < 97.97

Step-by-step explanation:

Step(i):-

Given sample mean (x⁻) = 95

standard deviation of the sample (s) = 6.6

Random sample size 'n' = 30

99% confidence interval for the mean score of all students.

step(ii):-

Degrees of freedom

ν =   n-1 = 30-1 =29

99% confidence interval for the mean score of all students.

( 95 - 2.966 , 95 + 2.966)

(92.03 , 97.97)

Final -

99% confidence interval for the mean score of all students.

92.03 < μ < 97.97

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are