Respuesta :
We have a variable x that is normally distributed with:
[tex]\begin{gathered} \mu=75 \\ \sigma=8 \end{gathered}[/tex]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:
[tex]P(ZThe value of z that solves the equation above cannot be found directly, it is solved by looking at a standard normal distribution table.Based on this, we find that the solution is z = 0.151 because from the normal table we see that:
[tex]P(Z<0.151)=0.56[/tex]Therefore, the percentile we are looking for is computed using the following formula:
[tex]\begin{gathered} P_{56}=\mu+z\cdot\sigma \\ P_{56}=75+0.151\cdot8 \\ P_{56}=76.208 \end{gathered}[/tex]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:
[tex]P(82\leq x\leq89)[/tex]Again, we must obtain the z-scores to solve this. The corresponding z-values needed to be computed are:
[tex]\begin{gathered} Z_{low}=\frac{x_1-\mu}{\sigma}=\frac{82-75}{8}=0.88 \\ Z_{up}=\frac{x_2-\mu}{\sigma}=\frac{89-75}{8}=1.75 \end{gathered}[/tex]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:
[tex]\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}[/tex]C. Finally, we must compute the following probability:
[tex]P(x\ge92)[/tex]The corresponding z-value needed to be computed is:
[tex]Z_{low}=\frac{x_1-\mu}{\sigma}=\frac{92-75}{8}=2.13[/tex]Again, because x follows a normal distribution, then the variable Zlow has a normal distribution and the probability is computed as:
[tex]\begin{gathered} P(x\ge92)=P(Z\ge Z_{low})_{} \\ P(x\ge92)=P(Z\ge2.13) \\ P(x\ge92)=0.0168 \end{gathered}[/tex]Summary
The results are:
A. 76.208
B. 0.1507
C. 0.0168
