Question

Assume that the game playthrough times for a newly released puzzle game has a mean of 49.8 minutes and a standard deviation of 4.2 minutes. a sample of size n=39 playthrough times is randomly selected from a gaming population. what is the probability that the sample mean is in between 50 minutes and 51 minutes? you may use a calculator or the portion of the z-table given below. round your answers to three decimal places where appropriate. z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.3 0.618 0.622 0.626 0.629 0.633 0.637 0.641 0.644 0.648 0.652 0.4 0.655 0.659 0.663 0.666 0.670 0.674 0.677 0.681 0.684 0.688 0.5 0.691 0.695 0.698 0.702 0.705 0.709 0.712 0.716 0.719 0.722 0.6 0.726 0.729 0.732 0.736 0.739 0.742 0.745 0.749 0.752 0.755 0.7 0.758 0.761 0.764 0.767 0.770 0.773 0.776 0.779 0.782 0.785 0.8 0.788 0.791 0.794 0.797 0.800 0.802 0.805 0.808 0.811 0.813 0.9 0.816 0.819 0.821 0.824 0.826 0.829 0.831 0.834 0.836 0.839 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

Answer 1

The probability that the sample mean is between 50 minutes and 51 minutes can be determined using the z-scores provided. Using the z-table, find the z-scores for 50 and 51 minutes, and then subtract the corresponding probabilities to find the desired probability.

Step-by-step explanation:

To find the z-scores for 50 and 51 minutes, we use the formula
`\(Z = ((X - \mu))/(\sigma)\),` where (X) is the value,
`\(\mu\)` is the mean, and
`\(\sigma\)` is the standard deviation. For 50 minutes,
`\(Z_(50) = ((50 - 49.8))/(4.2) \approx 0.048\),` and for 51 minutes,
`\(Z_(51) = ((51 - 49.8))/(4.2) \approx 0.286\).`

Next, we find the probabilities corresponding to these z-scores from the z-table. For
`/(Z_(50)\),` the probability is approximately 0.519, and for
`\(Z_(51)\),` the probability is approximately 0.611. Finally, we subtract
`\(P(Z < 0.286)\)`from
`\(P(Z < 0.048)\)` to find the probability that the sample mean is between 50 and 51 minutes. The result is approximately 0.092, rounded to three decimal places.