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The distribution of sample means for monthly employee income at a particular company for samples of 70 employees is​ normal, with a mean of ​$5971 and a standard deviation of ​$219 . You take a random sample of 70 employees from the company and find that their mean monthly income is ​$5 comma 751 . How many standard deviations is the sample mean from the mean of the distribution of sample​ means? Round to one decimal place.

User Hatter
by
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1 Answer

5 votes

Answer:

The value is
z = -8.4

Explanation:

From the question we are told that

The sample size is n = 70

The mean is
\mu = \$ 5971

The standard deviation is
\sigma = \$ 219

The sample mean is
\= x = \$ 5751

Generally the number of standard deviations the sample mean is from the mean of the distribution is mathematically represented as


z = (\= x - \mu )/( (\sigma)/(√(n) ) )

=>
z = ( 5751 -5971 )/( (219 )/(√( 70 ) ) )

=>
z = ( 5751 -5971 )/( (219 )/(√( 70 ) ) )

=>
z = -8.4

User Hoang Nguyen
by
5.9k points
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