Chapter 18, Problem 53E

(a)

To determine

To find the probability that this student’s IQ is at least $125$ points.

Answer to Problem 53E

  $0.734$ .

Explanation of Solution

In the question, suppose that IQs of East state university students can be described by a Normal model with mean $130$ and standard deviation $8$ points. Also suppose that IQs of students from West university can be described by a Normal model with mean $120$ and standard deviation $10$ . Thus, let us first find the z -score as:

  $\begin{array}{l}{z}_{125}=\frac{125-130}{8}\\ =-0.625\end{array}$

And the probability that this student’s IQ is at least $125$ points is calculated as:

  $\begin{array}{l}P(z\ge -0.625)=normalcdf(-0.625,E99,0,1)\\ =0.734\end{array}$

Thus, the probability that this student’s IQ is at least $125$ points is approximately $0.734$ .

(b)

To determine

To find the probability that the East State student’s IQ is at least $5$ points higher than the West State student’s IQ.

Answer to Problem 53E

  $0.652$ .

Explanation of Solution

In the question, suppose that IQs of East state university students can be described by a Normal model with mean $130$ and standard deviation $8$ points. Also suppose that IQs of students from West university can be described by a Normal model with mean $120$ and standard deviation $10$ . Thus, to find the probability we need to find the difference mean and standard deviation as:

  $\begin{array}{l}\mu =E(E-W)\\ =130-120\\ =10\\ \sigma =SD(E-W)\\ =\sqrt{Var(E)+Var(W)}\\ =\sqrt{64+100}\\ =12.806\end{array}$

Then the z -score will be as:

  $\begin{array}{l}{z}_5=\frac{5-10}{12.806}\\ =-0.390\end{array}$

Thus, the probability that the East State student’s IQ is at least $5$ points higher than the West State student’s IQ is calculated as:

  $\begin{array}{l}P(z\ge -0.390)=normalcdf(-0.390,E99,0,1)\\ =0.652\end{array}$

Thus, the probability that the East State student’s IQ is at least $5$ points higher than the West State student’s IQ is approximately $0.652$ .

(c)

To determine

To find the probability that this group’s average IQ is at least $125$ points.

Answer to Problem 53E

  $0.193$ .

Explanation of Solution

In the question, suppose that IQs of East state university students can be described by a Normal model with mean $130$ and standard deviation $8$ points. Also suppose that IQs of students from West university can be described by a Normal model with mean $120$ and standard deviation $10$ . Thus, we have,

  $\begin{array}{l}{\mu }_{\overline{y}}=120\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{10}{\sqrt{3}}\\ =5.774\end{array}$

Then the z -score will be as:

  $\begin{array}{l}{z}_{125}=\frac{125-120}{5.774}\\ =0.866\end{array}$

Thus, the probability that this group’s average IQ is at least $125$ points is calculated as:

  $\begin{array}{l}P(z\ge 0.866)=normalcdf(0.866,E99,0,1)\\ =0.193\end{array}$

Thus, the probability that this group’s average IQ is at least $125$ points is approximately $0.193$ .

(d)

To determine

To find out what is the probability that their average IQ is at least $5$ points higher than the average for the three West Stators.

Answer to Problem 53E

  $0.751$ .

Explanation of Solution

In the question, suppose that IQs of East state university students can be described by a Normal model with mean $130$ and standard deviation $8$ points. Also suppose that IQs of students from West university can be described by a Normal model with mean $120$ and standard deviation $10$ . Thus, we have,

  $\begin{array}{l}{\mu }_{\overline{x}}=130\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{8}{\sqrt{3}}\\ =4.619\end{array}$

And we have,

  $\begin{array}{l}\mu ={\mu }_{x-y}\\ =130-120\\ =10\\ \sigma =\sqrt{Var(E)+Var(W)}\\ =7.394\end{array}$

Thus, the z -score will be then,

  $\begin{array}{l}{z}_5=\frac{5-10}{7.394}\\ =-0.676\end{array}$

And the probability that their average IQ is at least $5$ points higher than the average for the three West Stators is calculated as:

  $\begin{array}{l}P(z\ge -0.676)=normalcdf(-0.676,E99,0,1)\\ =0.751\end{array}$

So, the probability that their average IQ is at least $5$ points higher than the average for the three West Stators is approximately $0.751$ .