Vending MachinesQuarters are now manufactured so that they

have a mean weight of 5.670 g and a standard deviation of 0.062 g, and their weights are normally distributed. A vending machine is configured to accept only those quarters that weigh between 5.550 g and 5.790 g.

a. If 1 randomly selected quarter is inserted into the vending machine, what is the probability that it will be accepted?

b. If 4 randomly selected quarters are inserted into the vending machine, what is the probability that their mean weight is between 5.550 g and 5.790 g?

c. If you own the vending machine, which result is more important: the result from part (a) or part (b)?

It is given that the weights of quarters are normally distributed with a mean equal to 5.670 grams and a standard deviation equal to 0.062 grams.

Only those quarters are accepted that weigh between 5.55 grams and 5.790 grams.

a.

Let X denote the weight of the quarters.

A quarter is accepted only if it has a weight between 5.55 grams and 5.790 grams.

The probability of selecting a quarter that will be accepted is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(5.550<x<5.790\right)=P\left(\frac{5.550-\mu }{\sigma }<\frac{x-\mu }{\sigma }<\frac{5.790-\mu }{\sigma }\right) \\ =P\left(\frac{5.550-5.670}{0.062}<z<\frac{5.790-5.670}{0.062}\right) \\ =P\left(-1.94<z<1.94\right) \end{gathered}$$

$$\begin{gathered} =P\left(z<1.94\right)-P\left(z<-1.94\right) \\ =0.9738-0.0262 \\ =0.9476 \end{gathered}$$

Therefore, the probability that a randomly selected quarter will be accepted is 0.9476.

b.

Let denote the sample mean weight of the quarters.

The sample mean weight follows a normal distribution with a mean equal to ${\mu }_{\overline{x}}=\mu$ and a standard deviation equal to ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The sample size is equal to n=4.

The probability that the sample mean of the 4 selected quarters is between 5.550 grams and 5.790 grams is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(5.550<\overline{x}<5.790\right)=P\left(\frac{5.550-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{5.790-\mu }{\frac{\sigma }{\sqrt{n}}}\right) \\ =P\left(\frac{5.550-5.670}{\frac{0.062}{\sqrt{4}}}<z<\frac{5.790-5.670}{\frac{0.062}{\sqrt{4}}}\right) \\ =P\left(-3.87<z<3.87\right) \end{gathered}$$

$$\begin{gathered} =P\left(z<3.87\right)-P\left(z<-3.87\right) \\ =0.9999-0.0001 \\ =0.9998 \end{gathered}$$

Therefore, the probability that the sample mean weight is between 5.550 grams and 5.790 grams is 0.9998.

c.

The vendor requires that the individual quarters meet the required specifications as the vending machine tests the individual weights of the quarters.

Thus, the vending machine does not account for the mean weight of a set of quarters.

Also, the probability from part (a) is less than the probability from part (b), which implies that there is a greater chance for the sample mean weight of a sample of quarters to meet the specification in comparison to the individual weight of a quarter to meet the requirements.

Therefore, the probability that a single sample quarter weight lies in the acceptable range (part (a)) is more important and appropriate as compared to the probability that the sample mean weight of 4 quarters lies in the acceptable range (part (b)).