Question

According to the latest financial reports from a sporting goods store, the mean sales per customer was $75 with a population standard deviation of $6. The store manager believes 39 randomly selected customers spent more per transaction.

What is the probability that the sample mean of sales per customer is between $76 and $77 dollars?

You may use a calculator or the portion of the z -table given below. Round your answer to two decimal places if necessary.

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
1.0	0.841	0.844	0.846	0.848	0.851	0.853	0.855	0.858	0.860	0.862
1.1	0.864	0.867	0.869	0.871	0.873	0.875	0.877	0.879	0.881	0.883
1.2	0.885	0.887	0.889	0.891	0.893	0.894	0.896	0.898	0.900	0.901
1.3	0.903	0.905	0.907	0.908	0.910	0.911	0.913	0.915	0.916	0.918
1.4	0.919	0.921	0.922	0.924	0.925	0.926	0.928	0.929	0.931	0.932
1.5	0.933	0.934	0.936	0.937	0.938	0.939	0.941	0.942	0.943	0.944
1.6	0.945	0.946	0.947	0.948	0.949	0.951	0.952	0.953	0.954	0.954
1.7	0.955	0.956	0.957	0.958	0.959	0.960	0.961	0.962	0.962	0.963
1.8	0.964	0.965	0.966	0.966	0.967	0.968	0.969	0.969	0.970	0.971
1.9	0.971	0.972	0.973	0.973	0.974	0.974	0.975	0.976	0.976	0.977
2.0	0.977	0.978	0.978	0.979	0.979	0.980	0.980	0.981	0.981	0.982
2.1	0.982	0.983	0.983	0.983	0.984	0.984	0.985	0.985	0.985	0.986
2.2	0.986	0.986	0.987	0.987	0.987	0.988	0.988	0.988	0.989	0.989

$\mu_{\overline{x}}=$ $

sigma sub line segment x is equal to $\sigma_{\overline{x}}=$

$

cap p times open paren 76 is less than or equal to line segment x comma line segment x is less than or equal to 77 close paren is equal to $P\left(76\le\overline{x}\le77\right)=$

Answer 1

Solution :

Given that ,

mean = = $75

standard deviation = = $6

n = 39

=

=  / n = 6 / 39=0.9608

P($76<   <$77 ) = P[(76 - 75) / 0.9608< ( - ) /   < (77 - 75) /0.9608 )]

= P( 1.04< Z <2.08 )

= P(Z < 2.08) - P(Z <1.04 )

Using z table,  

= 0.9812 -0.8508

=0.1304  

=0.13