Estimating demand for white bread. A bakery has determined that the number of loaves of its white bread demanded daily has a normal distribution with mean 7,200 loaves and standard deviation 300 loaves. Based on cost considerations, the company has decided that its best strategy is to produce a sufficient number of loaves so that it will fully supply demand on 94% of all days.

a. How many loaves of bread should the company produce?

b. Based on the production in part a, on what percentage of days will the company be left with more than 500 loaves of unsold bread?

A bakery has determined that the number of loaves of its white bread demanded daily has a normal distribution with mean 7,200 loaves and standard deviation 300 loaves.

Let x be the random variable that the number of the loaves of the white bread demanded daily.

Consider, $\mu -2\sigma$ and $\sigma =300$

The fully supply demand is on 94% of all days.

Consider,

$P(x\le x_0)=0.94$

From z score table, the value of z that corresponds to the probability is 1.56.

The number of loaves of bread the company should produce is obtained below:

$\mathit{z}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{0}}\mathbf{-}\mathbf{\mu }}{\mathbf{\sigma }}$

role="math" localid="1658221665216" $\begin{array}{c}1.56=\frac{x_0-7200}{300}\\ x_0=468+7200\\ =7668\end{array}$

Thus, the number of loaves of bread the company produced is 7668.

The company produces a total of 7668 loaves, then the company is left with more than 500 loaves is obtained if the demand is less than 7668 is$7668-500=7168$

The probability is obtained below:

role="math" localid="1658221647072" $\begin{array}{c}P(x<7168)=P(z<\frac{7168-7200}{300})\\ =P(z<\frac{-32}{300})\\ =P(z<-0.11)\end{array}$

$\begin{array}{l}=0.5-0.0438\\ =0.4562\end{array}$

Thus, the percentage of days that the company is left with more than 500 loaves is 45.62%.