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.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline$z$ & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\
\hline 1.0 & 0.841 & 0.844 & 0.846 & 0.848 & 0.851 & 0.853 & 0.855 & 0.858 & 0.860 & 0.862 \\
\hline 1.1 & 0.864 & 0.867 & 0.869 & 0.871 & 0.873 & 0.875 & 0.877 & 0.879 & 0.881 & 0.883 \\
\hline 1.2 & 0.885 & 0.887 & 0.889 & 0.891 & 0.893 & 0.894 & 0.896 & 0.898 & 0.900 & 0.901 \\
\hline 1.3 & 0.903 & 0.905 & 0.907 & 0.908 & 0.910 & 0.911 & 0.913 & 0.915 & 0.916 & 0.918 \\
\hline 1.4 & 0.919 & 0.921 & 0.922 & 0.924 & 0.925 & 0.926 & 0.928 & 0.929 & 0.931 & 0.932 \\
\hline 1.5 & 0.933 & 0.934 & 0.936 & 0.937 & 0.938 & 0.939 & 0.941 & 0.942 & 0.943 & 0.944 \\
\hline 1.6 & 0.945 & 0.946 & 0.947 & 0.948 & 0.949 & 0.951 & 0.952 & 0.953 & 0.954 & 0.954 \\
\hline 1.7 & 0.955 & 0.956 & 0.957 & 0.958 & 0.959 & 0.960 & 0.961 & 0.962 & 0.962 & 0.963 \\
\hline 1.8 & 0.964 & 0.965 & 0.966 & 0.966 & 0.967 & 0.968 & 0.969 & 0.969 & 0.970 & 0.971 \\
\hline 1.9 & 0.971 & 0.972 & 0.973 & 0.973 & 0.974 & 0.974 & 0.975 & 0.976 & 0.976 & 0.977 \\
\hline 2.0 & 0.977 & 0.978 & 0.978 & 0.979 & 0.979 & 0.980 & 0.980 & 0.981 & 0.981 & 0.982 \\
\hline 2.1 & 0.982 & 0.983 & 0.983 & 0.983 & 0.984 & 0.984 & 0.985 & 0.985 & 0.985 & 0.986 \\
\hline 2.2 & 0.986 & 0.986 & 0.987 & 0.987 & 0.987 & 0.988 & 0.988 & 0.988 & 0.989 & 0.989 \\
\hline
\end{tabular}
Answer
\(\boxed{0.131}\) is the probability that the sample mean of sales per customer is between $76 and $77.
Step 1 ：Calculate the standard error of the mean using the formula \(SE = \frac{\sigma}{\sqrt{n}}\). Substituting the given values, we get \(SE = \frac{6}{\sqrt{39}} \approx 0.96\).
Step 2 ：Calculate the z-scores for $76 and $77 using the formula \(Z = \frac{(X - \mu)}{SE}\). Substituting the given values, we get \(Z1 = \frac{(76 - 75)}{0.96} \approx 1.04\) and \(Z2 = \frac{(77 - 75)}{0.96} \approx 2.08\).
Step 3 ：Look up the z-scores in the z-table to find the corresponding probabilities. We find that \(P(Z1) = 0.850\) and \(P(Z2) = 0.981\).
Step 4 ：Calculate the probability that the sample mean is between $76 and $77 by subtracting the probabilities found in the previous step. We get \(P(76 < X < 77) = P(Z2) - P(Z1) = 0.981 - 0.850 = 0.131\).
Step 5 ：\(\boxed{0.131}\) is the probability that the sample mean of sales per customer is between $76 and $77.
