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

i understand 100% 10/10 hit or miss?

step-by-step explanation:

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2