Normally distributed sum of values

Mike's bowling scores are normally distributed with mean 110 and standard deviation 13, while Jack's scores are normally distributed with mean 135 and standard deviation 10. If Mike and Jack each bowl one game, find the probability that the total of their scores is above 225 (assuming their scores are independent). Enter your answer as a decimal and round to at least 3 correct decimal places.

I'm confused, how can I find the probability of their total score? I know that given the information I can find the variance but I'm confused as to that would help me with the final answer

• Let $X$ be Mike's score, and let $Y$ be Jack's. By the answer to your previous question, you know that $X+Y$ is normal and you can find the mean $\mu$ of $X+Y$, and the variance $\sigma^2$ of $X+Y$. (But of course $X+Y$ is not truly normal, bowling scores are integers.) The probability that a normal with mean $\mu$ is greater than $225$ is the probability that $Z\gt \frac{225-\mu}{\sigma}$, where $Z$ is standard normal. Three decimal places is ridiculous, you may be expected to make an (unjustified) continuity correction. – André Nicolas Apr 8 '14 at 6:19
• Should I be looking to calculate 1 - the probability that the standard normal is greater? – user130272 Apr 8 '14 at 6:25
• The mean is $245$, so $225-\mu=-20$, negative. So to use the standard tables, you will equivalently want to find the probability that $Z\lt \frac{20}{\sigma}$. (If you are using software, this step will likely not be necessary.) – André Nicolas Apr 8 '14 at 6:32
• @user130272 $1-\Pr\{Z>x\}=\Pr\{Z\le x\}$. How are you going to calculate the probability? Are you allowed to use statistical software? – Cm7F7Bb Apr 8 '14 at 6:33
• I'm learning how to code in python and yeah we're allowed to use other software. – user130272 Apr 8 '14 at 6:38

Hint: If $X\sim\mathcal N(\mu_X,\sigma_X^2)$ and $Y\sim\mathcal N(\mu_Y,\sigma_Y^2)$ are independent, then $$X+Y\sim\mathcal N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2).$$