Chapter 2.2, Problem 89E

To determine

Approximate standard deviation of weight for this species of cockroaches.

Answer to Problem 89E

Standard deviation,

  $\sigma \approx 4.6$

Explanation of Solution

Given information:

Weight, $x=55\text{grams}$

Mean, $\mu =50\text{grams}$

14% of the cockroaches weigh more than 55 grams.

Then

  $P\left(x>55\right)=14\%=0.14$

The total probability needs to be equal to 1.

Then

The probability of cockroaches weigh less than or equal to 55 grams:

  $\begin{array}{l}P\left(x\le 55\right)=1-P\left(x>55\right)\\ =1-0.14\\ =0.86\end{array}$

Now,

Find the z − score corresponding the probability of 0.86 in the normal probability table of the appendix.

Note that

The closest probability would be 0.8599 that lies in the row 1.0 and in the column .08 of the normal probability table.

Then

The corresponding z − score,

  $z=1.08$

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }$

Multiply both sides by standard deviation $\left(\sigma \right)$ :

  $z\sigma =x-\mu$

Divide both sides by z :

  $\sigma =\frac{x-\mu }{z}$

Substitute the values:

  $\sigma =\frac{x-\mu }{z}=\frac{55-50}{1.08}\approx 4.6$

Thus,

The standard deviation is approx. 4.6.