Question

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5%. Assume that a sample size of 40 people was surveyed from the population an infinite number of times.

95% of the sample mean occurs between:
A) 55.89%
B) 56.07%
C) 56.45%
and
A) 57.55%
B) 57.93%
C) 58.11%.

Answer 1

95% of the sample mean occurs between 56.45% and 57.55

Explanation:

I just did it on Edmentum

Answer 2

55.89 % < X < 58.11%

Explanation:

We have a Normal Curve.

P( L < Z < U ) = 95%

Two-tailed Confidence Interval.

L is above the 2.5%, so L = -1.96 from the chart

U is below the 2.5%, so U = 1.96

P( - 1.96 < Z < 1.96) = 95%

[(x - 57) / 3.5 ]* root(40) = -1.96

X = 55.915 Lower

(U - 57) / 3.5 *root(40) = 1.96

U - 57 = 1.96*3.5/root(40) = 1.08466

U = 58.084

so, from 55.9 < X < 58.1