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Q. 24

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The Practice of Statistics for AP

Found in: Page 803

Book edition 4th

Author(s) David Moore,Daren Starnes,Dan Yates

Pages 809 pages

ISBN 9781319113339

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Short Answer

Western lowland gorillas, whose main habitat is the central African continent, have a mean weight of $275 p o u n d s$ with a standard deviation of $40 p o u n d s$ . Capuchin monkeys, whose main habitat is Brazil and a few other parts of Latin America, have a mean weight of $6 p o u n d s$ with a standard deviation of $1.1 p o u n d s$ . Both weight distributions are approximately Normally distributed. If a particular western lowland gorilla is known to weigh $345 p o u n d s$ , approximately how much would a capuchin monkey have to weigh, in pounds, to have the same standardized weight as the lowland gorilla?
(a) $4.08$
(b) $7.27$
(c) $7.93$
(d) $8.20$
(e) There is not enough information to determine the weight of a capuchin monkey.

As the weight of a capuchin monkey is, $x \approx 7.93$

So, (c) option is the correct one.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

Mean weight of western lowland gorillas is, $a = 275 p o u n d s$

And its standard deviation is, $S . D_{1} = 40 p o u n d s$

And its weight is, $w = 345 p o u n d s$

Mean weight of Capuchin monkeys is, $b = 6 p o u n d s$

And its standard deviation is, $S . D_{2} = 1.1 p o u n d s$

Step 2: Simplify

We know that,

The z-score of gorilla is given as,

$z = \frac{w - a}{S . D_{1}} = \frac{345 - 275}{40} = 1.75$

As the standardized weight of capuchin monkey is same as that of gorilla. So, their z-scores must be equal to each other i.e. z-score of capuchin monkey is, $z = 1.75$

Now, as we know z- score is given as,

$z = \frac{x - b}{S . D_{2}} = 1.75 \frac{x - 6}{1.1} = 1.75 x - 6 = 1.75 \times 1.1 x - 6 = 1.925 x = 7.925 \approx 7.93$

Therefore, from above we get the correct option i.e. (c)

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