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Jeff's bowling scores are approximately normally distributed with mean 100 and standard deviation 11, while pam's scores are normally distributed with mean 105 and standard deviation 12. if jeff and pam each bowl one game, then assuming that their scores are independent random variables, approximate the probability that the total of their scores is above 190.

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Final answer:

The probability that the total score of Jeff and Pam's bowling games is above 190 is calculated by determining the combined mean and standard deviation of their scores, then finding the z-score for a total score of 190, and finally using a standard normal distribution to find the probability, which is approximately 82.12%.

Step-by-step explanation:

To find the probability that the total score of Jeff and Pam's bowling games is above 190, we first need to understand that when two independent normal random variables are added together, their means add and their variances add. Therefore, we can find the mean and variance of the total score.

Jeff's scores have a mean (μJeff) of 100 and a standard deviation (σJeff) of 11, so the variance is σJeff^2 = 121. Pam's scores have a mean (μPam) of 105 and a standard deviation (σPam) of 12, so the variance is σPam^2 = 144.

The mean of the total score, μtotal, is the sum of their means: μJeff + μPam = 100 + 105 = 205.

The variance of the total score, σtotal^2, is the sum of their variances: σJeff^2 + σPam^2 = 121 + 144 = 265. The standard deviation of the total score, σtotal, is the square root of the variance, which is approximately √265 ≈ 16.28.

Next, we need to find the z-score for a total score of 190. The z-score is given by (X - μtotal) / σtotal. For a score of 190, the z-score would be (190 - 205) / 16.28, which is approximately -0.921.

Finally, we look up this z-score in a standard normal distribution table or use a calculator to find that the probability of a z-score being above -0.921 is approximately 0.8212. Thus, the probability that the total of their scores is above 190 is approximately 0.8212 or 82.12%.

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