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 % and %.
Accepted Solution
A:
Answer:55.89 % < X < 58.11%Step-by-step 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 chartU is below the 2.5%, so U = 1.96P( - 1.96 < Z < 1.96) = 95%[(x - 57) / 3.5 ]* root(40) = -1.96X = 55.915 Lower(U - 57) / 3.5 *root(40) = 1.96U - 57 = 1.96*3.5/root(40) = 1.08466U = 58.084so, from 55.9 < X < 58.1