A random sample of 25 statistics examinations was selected. The average score in the sample was 76 with a variance of 144. Assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is .

99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 76

variance = 144

sd = sqrt(variance) = sqrt(144) = 12

n = 25

degree of freedom (df) = n - 1 = 25 - 1 = 24

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 24 df and 1% significance level is 2.797

E = t×sd/√n = 2.797×12/√25 = 6.7128

Lower limit = mean - E = 76 - 6.7128 = 69.2872

Upper limit = mean + E = 76 + 6.7128 = 82.7128

99% confidence interval is (69.2872, 82.7128)

i would need to see the dot plot : )

step-by-step explanation: