



 
Hi Alysia, You first need to convert the normal distribution with random variable $X$ with mean $\mu = 66$ and standard distribution $\sigma = 12$ to the standard normal distribution with random variable $Z.$ The conversion is \[Z = \frac{X  \mu}{\sigma} = \frac{X  66}{12} \] Now you need to use the standard distribution table to find the number $Z_{0.08}$ so that the area under the normal curve and to the right of $Z_{0.08}$ is $0.08,$ as illustrated in the diagram below. Set $Z = Z_{0.08}$ and solve for $X.$ Penny  


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