Question

QUESTION 17.1 POINT 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 1.00.841 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 0.8870.889 0.891 0.8930.894 0.896 0.898 0.900 0.901 1.3 0.9030.905 0.9070.908 0.910 0.911 0.9130.915 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 0.9390.941 0.942 0.943 0.944 1.60.945 0.946 0.947 0.948 0.949 0.951 0.9520.9530.954 0.954 1.7 0.955 0.956 0.957 0.958 0.959 0.9600.961 0.962 0.962 0.963 1.8 0.964 0.965 0.9660.966 0.967 0.968 0.969 0.9690.970 0.971 1.9 0.971 0.972 0.973 0.9730.974 0.9740.975 0.976 0.976 0.977 2.0 |0.9770.978 0.978 0.979 0.979 0.980 0.980 0.981 0.981 0.982 2.1 0.982 0.9830.9830.9830.984 0.9840.985 0.985 0.985 0.986 2.210.986 0.986 0.987 0.987 0.987 0.988 0.988 0.988 0.989 0.989

Answer 1

6x = 0 Given, mean =H = $ 75 populatim stondass deviation av =$6 Sample size=n=39 My = "=$1.75 6 کاره 139 y = $10.96

P(7656 577)-p/16-Mx 77-Mx < X-M of 8 Accoeling to central limit thecrom. 79-75 0.96 0.96 p/76-75 52= = P(1.043232.08) =PLZE 2.08) - PIZ< 1.04) using z table, we get = 0.981-0.851 P1765 x srila 0.13