Question

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.

Short answer

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

So, (c) option is the correct one.

Step-by-step solution

Given information

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,$S.D_2=1.1pounds$

Simplify

We know that,

The z-score of gorilla is given as,

$$\begin{gathered} z=\frac{w-a}{S.D_1} \\ =\frac{345-275}{40} \\ =1.75 \end{gathered}$$

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,

$$\begin{gathered} z=\frac{x-b}{S.D_2}=1.75 \\ \frac{x-6}{1.1}=1.75 \\ x-6=1.75\times 1.1 \\ x-6=1.925 \\ x=7.925\approx 7.93 \end{gathered}$$

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