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Q. 24

Expert-verifiedFound in: Page 803

Book edition
4th

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

Pages
809 pages

ISBN
9781319113339

Western lowland gorillas, whose main habitat is the central African continent, have a mean weight of $275pounds$ with a standard deviation of $40pounds$. Capuchin monkeys, whose main habitat is Brazil and a few other parts of Latin America, have a mean weight of $6pounds$ with a standard deviation of $1.1pounds$. Both weight distributions are approximately Normally distributed. If a particular western lowland gorilla is known to weigh $345pounds$, 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.

Mean weight of western lowland gorillas is, $a=275pounds$

And its standard deviation is, $S.{D}_{1}=40pounds$

And its weight is, $w=345pounds$

Mean weight of Capuchin monkeys is, $b=6pounds$

And its standard deviation is, $S.{D}_{2}=1.1pounds$

We know that,

The z-score of gorilla is given as,

$z=\frac{w-a}{S.{D}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{345-275}{40}\phantom{\rule{0ex}{0ex}}=1.75$

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,

$z=\frac{x-b}{S.{D}_{2}}=1.75\phantom{\rule{0ex}{0ex}}\frac{x-6}{1.1}=1.75\phantom{\rule{0ex}{0ex}}x-6=1.75\times 1.1\phantom{\rule{0ex}{0ex}}x-6=1.925\phantom{\rule{0ex}{0ex}}x=7.925\approx 7.93$

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

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