Question

Random variable is normally distributed with mean 75 and standard deviation 8.a. Find the 56th percentile for this random variable.b. Find the proportion of the values for random variable between 82 and 89.c. Find the probability that the randomly selected random variable is greater than 92.

Answer 1

We have a variable x that is normally distributed with:

`\begin{gathered} \mu=75 \\ \sigma=8 \end{gathered}`

A. Using the data above we must find the 56th-percentile for the random variable x.

First, we need to find the z-score associated with this percentile. How we do that? We must find the value of z that solves the following equation:

Therefore, the percentile we are looking for is computed using the following formula:

`\begin{gathered} P_(56)=\mu+z\cdot\sigma \\ P_(56)=75+0.151\cdot8 \\ P_(56)=76.208 \end{gathered}`

B. We must find the proportion of the values for the random variable x between 82 and 89.

In mathematical terms, in this case, we must compute the following probability:

`P(82\leq x\leq89)`

Again, we must obtain the z-scores to solve this. The corresponding z-values needed to be computed are:

`\begin{gathered} Z_(low)=(x_1-\mu)/(\sigma)=(82-75)/(8)=0.88 \\ Z_(up)=(x_2-\mu)/(\sigma)=(89-75)/(8)=1.75 \end{gathered}`

Now, because the variable x is a normal distribution, then the variables Zlow and Zup have a normal distribution. Therefore, the probability is computed in the following way:

`\begin{gathered} P(82\leq x\leq89)=P(Z_{\text{low}}\leq z\leq Z_(up)) \\ P(82\leq x\leq89)=P(0.88\leq z\leq1.75) \\ P(82\leq x\leq89)=P(Z\leq1.75)-P(Z\leq0.88) \\ P(82\leq x\leq89)=0.9599-0.8092 \\ P(82\leq x\leq89)=0.1507 \end{gathered}`

C. Finally, we must compute the following probability:

`P(x\ge92)`

The corresponding z-value needed to be computed is:

`Z_(low)=(x_1-\mu)/(\sigma)=(92-75)/(8)=2.13`

Again, because x follows a normal distribution, then the variable Zlow has a normal distribution and the probability is computed as:

`\begin{gathered} P(x\ge92)=P(Z\ge Z_(low))_{} \\ P(x\ge92)=P(Z\ge2.13) \\ P(x\ge92)=0.0168 \end{gathered}`

Summary

The results are:

A. 76.208

B. 0.1507

C. 0.0168