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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%.

Respuesta :

mysticchacha

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 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

Answer:

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

Step-by-step explanation:

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