Question

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.

Answer 1

Answer:

The value is  $z = -8.4$

Step-by-step 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

         $z = \frac{\= x - \mu }{ \frac{\sigma}{\sqrt{n} } }$

=>      $z = \frac{ 5751 -5971 }{ \frac{219 }{\sqrt{ 70 } } }$

=>      $z = \frac{ 5751 -5971 }{ \frac{219 }{\sqrt{ 70 } } }$    

=>      $z = -8.4$