91Ó°ÊÓ

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, $\mathrm{μ}−2\mathrm{σ}$ and $\mathrm{σ}=300$

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

Consider,

$P(x≤x_0)=0.94$

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:

$\mathit{z}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{0}}\mathbf{−}\mathbf{μ}}{\mathbf{σ}}$

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%.