Question: The owner of a minor league baseball team pays the nine starting players, the manager, and two…

The owner of a minor league baseball team pays the nine starting players, the manager, and two coaches bonuses, which are certain percentages of their regular salaries each time the team wins a game. The amounts paid are listed below. How many of the bonus amounts are within one standard deviation of the mean?

$80 $105 $145 $190 $200 $205 $220 $225 $230 $310 $330 $335

Solution

To solve this problem, we need to find the number of bonus amounts that are within one standard deviation of the mean. We will follow these steps: First, calculate the mean of the bonuses. The bonuses given are: \[ 80, 105, 145, 190, 200, 205, 220, 225, 230, 310, 330, 335 \] Number of bonuses = 12 Mean: \[ \text{Mean} = \frac{80 + 105 + 145 + 190 + 200 + 205 + 220 + 225 + 230 + 310 + 330 + 335}{12} \] \[ \text{Mean} = \frac{2575}{12} = 214.58 \] Next, calculate the standard deviation. First, calculate the variance: \[ \text{Variance} = \frac{(80-214.58)^2 + (105-214.58)^2 + \ldots + (335-214.58)^2}{12} \] \[ \text{Variance} = \frac{18061.67}{12} = 1505.14 \] Standard deviation is the square root of the variance: \[ \text{Standard Deviation} = \sqrt{1505.14} \approx 38.79 \] Determine the range for one standard deviation around the mean: \[ \text{Mean} - \text{Standard Deviation} = 214.58 - 38.79 = 175.79 \] \[ \text{Mean} + \text{Standard Deviation} = 214.58 + 38.79 = 253.37 \] Count the number of bonuses within this range (175.79 to 253.37): - 190 - 200 - 205 - 220 - 225 - 230 There are 6 bonuses within one standard deviation of the mean.