Chapter 2, Problem R2.7RE

(a)

To determine

Percent of time the restaurant’s goal will happen.

Answer to Problem R2.7RE

Between 1 and 1.2 ounces of the ketchup has been put on a burger 70.56% of time.

Explanation of Solution

Given information:

  $x=1\text{ounce}\text{and}1.2\text{ounces}$

Mean, $\mu =1.05\text{ounces}$

Standard deviation,

  $\sigma =0.08\text{ounce}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{1-1.05}{0.08}\approx -0.63$

And

  $z=\frac{x-\mu }{\sigma }=\frac{1.2-1.05}{0.08}\approx 1.88$

Use normal probability table in the appendix, to find the corresponding probability.

  $\begin{array}{l}P\left(1<x<1.2\right)=P\left(-0.63<z<1.88\right)\\ =P\left(z<1.88\right)-P\left(z<-0.63\right)\\ =0.9699-0.2643\\ =0.7056\\ =70.56\%\end{array}$

Therefore,

Between 1 and 1.2 ounces of the ketchup has been put on a burger 70.56% of time.

(b)

To determine

Reduction in machine’s standard deviation to ensure at least 99% of the restaurant’s burgers have between 1 and 1.2 ounces ketchup on them.

Answer to Problem R2.7RE

We need to reduce the standard deviation by 0.04124 ounces.

Explanation of Solution

Given information:

  $x=1\text{ounce}\text{and}1.2\text{ounces}$

Mean, $\mu =1.05\text{ounces}$

Calculations:

We have

At least 99% of the burgers need between 1 and 1.2 ounces of ketchup.

  $P\left(1<x<1.2\right)=99\%=0.99$

Note that

Mean is exactly in the middle i.e. 1.1.

Thus,

99% corresponds with the middle 99%.

Thus,

  $\frac{100\%-99\%}{2}=0.5\%$

There are about 0.05% burgers on either side of the interval.

This implies

There are 0.5% of the burgers with less than 1 ounce of ketchup.

However,

  $99\%+0.5\%=99.5\%$

There are 99.5% of the burgers with less than 1.2 ounces of ketchup.

In the normal probability table, z − scores corresponds with a probability of 0.005 (0.5%) and 0.995 (99.5%), or the probability closest.

  $z=\pm 2.58$

Now,

z − score:

  $z=\frac{x-\mu }{\sigma }$

Multiply both sides by $\sigma$ :

  $z\sigma =x-\mu$

Divide each side by z − score ( z ):

  $\sigma =\frac{x-\mu }{z}$

Substitute the known values and substitute:

Note that

  $z=-2.58$ corresponds with $x=1$

And

  $z=2.5$ corresponds with $x=1.2$

Then

  $\sigma =\frac{x-\mu }{z}=\frac{1-1.1}{-2.58}=\frac{-0.1}{-2.58}=\frac{0.1}{2.58}\approx 0.03876$

The standard deviation is approx. is 0.03876 ounces.

The original standard deviation is 0.08 ounces.

Subtracting standard deviation calculated from the original standard deviation, we get 0.04124 ounces.

Thus,

We need to reduce the standard deviation by 0.04124 ounces.