Question

Ketchup A fast-food restaurant has just installed a new automatic ketchup dispenser for use in preparing its burgers. The amount of ketchup dispensed by the machine can be modeled by a Normal distribution with mean $1.05$ ounces and a standard deviation $0.08$ ounces.

a. If the restaurant’s goal is to put between $1$ and $1.2$ ounces of ketchup on each burger, about what percent of the time will this happen?

b. Suppose that the manager adjusts the machine’s settings so that the mean amount of ketchup dispenser is $1.1$ ounces. How much does the machine’s standard deviation have to be reduced to ensure that at least $99\%$ of the restaurant’s burgers have between $1$ and $1.2$ ounces of ketchup on them?

Short answer

Part (a) Between $1$and $1.2$ounces of the ketchup has been put on a burger $70.56\%$of time.

Part (b) We need to reduce the standard deviation by $0.04124$ ounces.

Step-by-step solution

Part (a) Step 1: Given information

$x=1ounceand1.2ounces$

Mean,$\mu =1.05ounces$

Standard deviation,$\sigma =0.08ounce$

Part (a) Step 2: Concept

The formula used:$z=\frac{x-\mu }{\sigma }$

Part (a) Step 3: Calculation

Calculate the $z-$score,

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

And

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

To find the equivalent probability, use the normal probability table in the appendix.

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

Therefore,

Between $1$ and $1.2$ ounces of the ketchup has been put on a burger data-custom-editor="chemistry" $70.56\%$ of time.

Part (b) Step 1: Calculation

We have

Between $1$and $1.2$ounces of ketchup is required for at least $99$percent of the burgers.

$P(1<x<1.2)=99\%=0.99$

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,

$99\%+0.5\%=99.5\%$

$99.5$percent of the burgers have less than $1.2$ounces of ketchup on them.

The z scores in the normal probability table correspond to a probability of $0.005(0.5$percent) to $0.995(99.5\%)$or the probability nearest.

$z=\pm 2.58$

Now,

$z-$score:

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

Multiply both sides by $\sigma$:

Divide each side by $z-$score $(z)$:

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

Substitute the following values for the known values:

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 around $0.03876$ ounces.

The original standard deviation is $0.08$ ounces.

We get $0.04124$ ounces by subtracting the estimated standard deviation from the original standard deviation.

Thus,

We need to reduce the standard deviation by $0.04124$ ounces.