Two students are throwing water balloons at a target. Accuracy is measured as how close the balloon is from the
center of the target. Tory's distances from the center of the target are approximately Normally distributed with a
mean of 123 mm and a standard deviation of 31 mm. Adam's distances from the center is approximately Normally
distributed with a mean of 108 mm and a standard deviation of 47 mm. If 5 attempts for Tory and 3 attempts for
Adam are randomly selected, what is the probability that the mean distance from the center of the target for Tory is
more than for Adam?
O 0.3113
O 0.4923
O 0.6887
O 0.7427

Tory's distances from the center of the target are normally distributed with a mean of 123mm and a standard deviation 31mm. Number of attempts for Tory is 5.

Adam's distances from the center of the target are with a mean of 108mm and a standard deviation 47mm. Number of attempts for Adam is 3.

In order to find the probability of that mean distance from the center of the target for Tory is more than for Adam.

Normal distribution z = (x-μ)/σ x ~ (μ,σ)

When x = 108

z = (123 - 108)/47 [Adam's distances from the center of the target are normally distributed with a mean of 108mm and a standard deviation of Adam is 31mm.]

= 0.3191

P(x > 108) = P(z > - 0.3191)

= 1 - 0.3191 [since the probability of the normal curve is 1 and to find the area under tory mean is greater than the Adam mean only if we subtract 1 from z value]

= 0.6887

Hence, Using normal distribution, the probability of that mean distance from the center of the target for Tory is more than for Adam is 0.6887.

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