91Ó°ÊÓ

Given in the question that, Sam has determined that the weights of unpeeled bananas from his local store have a mean of$116grams$with a standard deviation of $9grams$.

if the distribution of weight is approximately Normal, to the nearest gram, We need to find the heaviest $30\%$of the bananas weigh is at least how much.

This is a given,

$\mathrm{μ}=116$

$\mathrm{σ}=9$

The cutoff value for the heaviest $30$percent weighs the same as the lower $100\%-30\%=70\%$weighs.

Now we'll look for the z-score that corresponds to a likelihood of $70\%$in the normal probability. As a result, the zscore is:

$z=0.50+0.02$

$=0.52$

As a result,

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

$=\frac{x-116}{9}$

role="math" localid="1654188929717" $⇒\frac{x-116}{9}=0.52$

$⇒x-116=0.52\left(9\right)$

$⇒x=116+0.52\left(9\right)$

$⇒x=120.68$

As a result, the heaviest $30\%$of these bananas weigh at least $120.68\mathrm{g}$, or roughly$121\mathrm{g}$. As a result, option (b) is the proper choice.