Question

Sam has determined that the weights of unpeeled bananas from his local store have a mean of$116grams$ with a standard deviation of $9grams$. Assuming that the distribution of weight is approximately Normal, to the nearest gram, the heaviest $30\%$of these bananas weigh at least how much?

a.$107\mathrm{g}$

b.$121\mathrm{g}$

C.$111\mathrm{g}$

d.$125\mathrm{g}$

e.$116\mathrm{g}$

Short answer

The correct option is

$b.121\mathrm{g}$

Step-by-step solution

Given information

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.

Explanation

This is a given,

$\mu =116$

$\sigma =9$

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,

$z=\frac{x-\mu }{\sigma }$

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

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

$\Rightarrow x-116=0.52\left(9\right)$

$\Rightarrow x=116+0.52\left(9\right)$

$\Rightarrow x=120.68$

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.