91Ó°ÊÓ

$68-95-99.7$ rule

Approximately $68\%$ of the observations are within $1$ standard deviation of the mean.

Approximately $95\%$ of the observations are within $2$ standard deviations of the mean.

Approximately $99.7\%$ of the observations are within $3$ standard deviations of the mean.

Mean $=80$

Standard deviation $=2$

This means that $78$is $1$standard deviation below the mean $(\mathrm{μ}-\mathrm{σ}=80-2=78)$.

So, in order to determine approximately $68\%$of the data values, we use the $68-95-99.7$rule.

Because there are $100\%$of the data values in total, we expect $32\%$to exceed 1 standard deviation from the mean, since $100\%-68\%=32\%$of the data values are in excess of$1$.

In view of the symmetry of the normal distribution, the expectation is that exactly half of the $32\%$(that is,$16\%$) will be more than $1$standard deviation below the mean.

Consequently, $16\%$ of the cockroaches will weigh less than $78$ grams.

Therefore, option (c) is the correct answer.