Question

Jeff's bowling scores are approximately normally distributed with mean 100 and standard deviation 21, while Pam's scores are normally distributed with mean 155 and standard deviation 22. If Jeff and Pam each bowl one game, then assuming that their scores are independent random variables, approximate the probability that

a. the total of their scores is above 260.
b. Jeffs score is higher.

Answer 1

a. 0.4364 = 43.64% probability that the total of their scores is above 260.

b. 0.0351 = 3.51% probability that Jeff's score is higher.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Sum/Subtraction between normal variables:

When two normal variables are added/subtracted, the mean is the sum/subtraction of the means, while the standard deviation is the square root of the sum of the variances.

a. the total of their scores is above 260.

Mean is the sum of the means, so:

`\mu = 100 + 155 = 255`

Standard deviation is the square root of the sum of the variances. So

`\sigma = √(21^2+22^2) = 30.4`

The probability is 1 subtracted by the p-value of Z when X = 260. So

`Z = (X - \mu)/(\sigma)`

`Z = (260 - 255)/(30.4)`

`Z = 0.16`

`Z = 0.16` has a p-value of 0.5636

1 - 0.5636 = 0.4364

0.4364 = 43.64% probability that the total of their scores is above 260.

b. Jeff's score is higher.

Jeff's mean is of 100, while Pam's is of 155. The probability of Jeff's score being higher is the probability that the subtraction is above 0. The mean is:

`\mu = 100 - 155 = -55`

The probability is 1 subtracted by the p-value of Z when X = 0. So

`Z = (X - \mu)/(\sigma)`

`Z = (0 - (-55))/(30.4)`

`Z = 1.81`

`Z = 1.81` has a p-value of 0.9649.

1 - 0.9649 = 0.0351

0.0351 = 3.51% probability that Jeff's score is higher.