A different species of cockroach has weights that are approximately Normally distributed with a mean of $50$ grams. After measuring the weights of many of these cockroaches, a lab assistant reports that $14\%$ of the cockroaches weigh more than $55$ grams. Based on this report, what is the approximate standard deviation of weight for this species of cockroaches?

$$\begin{gathered} a.4.6 \\ b.5.0 \\ c.6.2 \\ d.14.0 \end{gathered}$$

e. Cannot determine without more information.

Weight,$x=55grams$

Mean,$\mu =50grams$

The formula used:$z=\frac{x-\mu }{\sigma }$

14% of the cockroaches weigh more than 55 grams.

Then

$P(x>55)=14\%=0.14$

The total probability needs to be equal to 1.

Then

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

$$\begin{gathered} P(x\le 55)=1-P(x>55) \\ =1-0.14 \\ =0.86 \end{gathered}$$

Now,

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

Note that

The closest probability would be 0.8599 which lies in row 1.0 and in 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 the standard deviation $\left(\sigma \right)$:

$z\sigma =x-\mu$

Divide both sides by 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.