The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 2.2, Problem 54E

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 54E

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

Explanation of Solution

Given:

  μ=110 and σ=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 μ=110 and σ=25 .

State: Let xbe the IQ scores of students. The variable x has a Normal distribution with μ=110 and σ=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:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 54E , additional homework tip  1

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

  z=15011025=1.625

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 95th percentile.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

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 μ=110 and σ=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:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 54E , additional homework tip  2

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:

  z=12511025=0.6

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

(c)

Expert Solution
Check Mark

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 μ=110 and σ=25 . We want to find the score such that 98% of people aged 20-34 have an IQ score that is smaller.

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

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 54E , additional homework tip  3

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

  2.05=x11025x=161.25

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.

Chapter 2 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 2.1 - Prob. 4.1CYUCh. 2.1 - Prob. 4.2CYUCh. 2.1 - Prob. 4.3CYUCh. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10ECh. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.2 - Prob. 1.1CYUCh. 2.2 - Prob. 1.2CYUCh. 2.2 - Prob. 1.3CYUCh. 2.2 - Prob. 2.1CYUCh. 2.2 - Prob. 2.2CYUCh. 2.2 - Prob. 2.3CYUCh. 2.2 - Prob. 2.4CYUCh. 2.2 - Prob. 2.5CYUCh. 2.2 - Prob. 3.1CYUCh. 2.2 - Prob. 3.2CYUCh. 2.2 - Prob. 3.3CYUCh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.2 - Prob. 44ECh. 2.2 - Prob. 45ECh. 2.2 - Prob. 46ECh. 2.2 - Prob. 47ECh. 2.2 - Prob. 48ECh. 2.2 - Prob. 49ECh. 2.2 - Prob. 50ECh. 2.2 - Prob. 51ECh. 2.2 - Prob. 52ECh. 2.2 - Prob. 53ECh. 2.2 - Prob. 54ECh. 2.2 - Prob. 55ECh. 2.2 - Prob. 56ECh. 2.2 - Prob. 57ECh. 2.2 - Prob. 58ECh. 2.2 - Prob. 59ECh. 2.2 - Prob. 60ECh. 2.2 - Prob. 61ECh. 2.2 - Prob. 62ECh. 2.2 - Prob. 63ECh. 2.2 - Prob. 64ECh. 2.2 - Prob. 65ECh. 2.2 - Prob. 66ECh. 2.2 - Prob. 67ECh. 2.2 - Prob. 68ECh. 2.2 - Prob. 69ECh. 2.2 - Prob. 70ECh. 2.2 - Prob. 71ECh. 2.2 - Prob. 72ECh. 2.2 - Prob. 73ECh. 2.2 - Prob. 74ECh. 2.2 - Prob. 75ECh. 2.2 - Prob. 76ECh. 2 - Prob. 1CRECh. 2 - Prob. 2CRECh. 2 - Prob. 3CRECh. 2 - Prob. 4CRECh. 2 - Prob. 5CRECh. 2 - Prob. 6CRECh. 2 - Prob. 7CRECh. 2 - Prob. 8CRECh. 2 - Prob. 9CRECh. 2 - Prob. 10CRECh. 2 - Prob. 11CRECh. 2 - Prob. 12CRECh. 2 - Prob. 1PTCh. 2 - Prob. 2PTCh. 2 - Prob. 3PTCh. 2 - Prob. 4PTCh. 2 - Prob. 5PTCh. 2 - Prob. 6PTCh. 2 - Prob. 7PTCh. 2 - Prob. 8PTCh. 2 - Prob. 9PTCh. 2 - Prob. 10PTCh. 2 - Prob. 11PTCh. 2 - Prob. 12PTCh. 2 - Prob. 13PT
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