We draw a random sample of size 25 from a normal population with variance 2.4. If the sample mean is 12.5, what is a 99% confidence interval for the population mean?

Step-by-step explanation:

Confidence interval for the population mean is expressed by the formula;

CI = xbar ± Z(S/√n) where;

xbar is the sample mean = 12.5

Z is the z score at 99% confidence = 2.576

S is the standard deviation = √variance

S = √2.4 = 1.5492

n is the sample size = 25

Substituting the given values into the formula given above,

CI = 12.5 ± 2.576(1.5492/√25)

CI = 12.5 ± 2.576(0.30984)

CI = 12.5 ± 0.7981

CI = (12.5-0.7981, 12.5+0.7981)

CI = (11.2019, 12.7981)

Hence the 99% confidence interval for the population mean is 11.2≤12.8 (to 1 decimal place)

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)