Chapter 2.2, Problem 54E

(a)

To determine

To find: the percentile where the IQ score is of 150

Answer to Problem 54E

An IQ score of 150 is approximately at the 95^th percentile.

Explanation of Solution

Given:

  $\mu =110$ and $\sigma =25$ .

Calculation:

Scores on the Wechsler adult Intelligence Scale (a standard IQ test) for the 20 to 34 age groupare approximately Normal with mean 110 and standard deviation 25.

That is $\mu =110$ and $\sigma =25$ .

State: Let xbe the IQ scores of students. The variable x has a Normal distribution with $\mu =110$ and $\sigma =25$ .We want the proportion of IQ scores that are less than 150.

Plan: The proportion of IQ scores less than 150 is shown in the graph below:

  

Do: For x =150, the corresponding z value is calculated as follows:

  $\begin{array}{l}z=\frac{150-110}{25}\\ =1.625\end{array}$

Using standard normal table, we can see that the proportion of observations less than 1.6 is 0.9452 or approximately 94.5%.

Conclude: An IQ score of 150 is approximately at the 95^th percentile.

Conclusion:

Therefore, An IQ score of 150 is approximately at the 95^th percentile.

(b)

To determine

To find: the percent of people aged 20 to 34 have IQs between 125 and 150

Answer to Problem 54E

Approximately 22% of people aged 20-34 have IQ scores between 12 and 150.

Explanation of Solution

Calculation:

State: Let x be the IQ scores of students. The variable x has a Normal distribution with $\mu =110$ and $\sigma =25$ .We want the proportion of IQ scores that are between 125 and 150.

Plan: The proportion of IQ scores between 125 and 150 is shown in the graph below:

  

Do: From part (a) we have that for x = 150, the corresponding z - value = 1.6. For x =125, the corresponding z -value is calculated as follows:

  $\begin{array}{l}z=\frac{125-110}{25}\\ =0.6\end{array}$

So the proportion of observations between 0.6 and 1.6, is equal to the proportion of observations between 125 and 150.

Using standard normal table, the area between 0.6 and 1.6 is 0.9452 - 0.7257 =0.2195 or about 22%.

Conclude: Approximately 22% of people aged 20-34 have IQ scores between 12 and 150.

Conclusion:

Therefore, 22% of people aged 20-34 have IQ scores between 12 and 150.

(c)

To determine

To find: the scored to earn to qualify for MENSA membership

Answer to Problem 54E

In order to qualify for MENSA membership a person must score 162 or higher.

Explanation of Solution

Calculation:

State: Let xbe the IQ scores of students. The variable x has a Normal distribution with $\mu =110$ and $\sigma =25$ . We want to find the score such that 98% of people aged 20-34 have an IQ score that is smaller.

Plan: The 98^th percentile of the IQ scores is shown in the graph below:

  

Do: From standard normal table, the 98" percentile for the standard Normal distribution is closest to 2.05. Therefore, the 98^th percentile for the IQ scores can be found by solving the equation:

  $\begin{array}{l}2.05=\frac{x-110}{25}\\ x=161.25\end{array}$

Conclude: In order to qualify for MENSA membership a person must score 162 or higher.

Conclusion:

Therefore, in order to qualify for MENSA membership a person must score 162 or higher.