Question

Quarters After 1964, quarters were manufactured so that their weights have a mean of5.67 g and a standard deviation of 0.06 g. Some vending machines are designed so that you canadjust the weights of quarters that are accepted. If many counterfeit coins are found, you cannarrow the range of acceptable weights with the effect that most counterfeit coins are rejectedalong with some legitimate quarters.

a. If you adjust your vending machines to accept weights between 5.60 g and 5.74 g, what percentage of legal quarters are rejected? Is that percentage too high?

b. If you adjust vending machines to accept all legal quarters except those with weights in the top 2.5% and the bottom 2.5%, what are the limits of the weights that are accepted?

Short answer

a. The percent of legal quarters are rejected is24.2%.Yes, the percentage is high.

b. 5.55g and 5.79gare the limits of weights that are accepted.

Step-by-step solution

Given information

Quarters were manufactured so that the weights were normally distributed with a mean of 5.67g and a standard deviation of 0.06g.

Step 2:Find the z score corresponding to the specified value.

a.

Let X be the weights of legal quarters.

$$\begin{gathered} X~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(5.67,{0.06}^2\right) \end{gathered}$$

The acceptable range of weights for vending machines is 5.60 g to 5.74 g.

The percentage of legal quarters accepted by vending machines is computed using the z score.

For $x=5.74$, the z-score is given by:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{5.74-5.67}{0.06} \\ =1.17 \end{gathered}$$

For$x=5.60$, the z-score is given by:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{5.6-5.67}{0.06} \\ =-1.17 \end{gathered}$$

Find the probabilities corresponding to the value of z score

Referring to the standard normal table for positive z score, the cumulative probability of 1.17 is obtained from the cell intersection for rows 1.1 and the column value 0.07,which implies that the z score is 0.8790.

Referring to the standard normal table for negative z score, the cumulative probability of 1.17 is obtained from the cell intersection for rows -1.1 and the column value 0.07,which implies that the z score is 0.1210.

Thus,

$$\begin{gathered} P\left(Z<-1.17\right)=0.1210 \\ P\left(Z<1.17\right)=0.8790 \end{gathered}$$

Find the probability

The probability that a random quarter weighs between 5.60g and 5.74g is:

$$\begin{gathered} P\left(-1.17<Z<1.17\right)=\text{Area}\text{less}\text{than}1.17-\text{Area}\text{less}\text{than}-1.17 \\ =P\left(Z<1.17\right)-P\left(Z<-1.17\right) \\ =0.8790-0.1210 \\ =0.7580 \end{gathered}$$

By expressing the result as a percentage, it is concluded that 75.8% of the legal quarters are accepted.

Calculate the percentage of legal quarters that are rejected

Rejection of the legal quarter is a complementary event of accepting the quarter.

Thus,

$$\begin{gathered} \text{Probability of legal quarters are rejected}=\text{1}-\text{Probability of legal quarters are accepted} \\ =1-0.758 \\ =0.242 \end{gathered}$$

By expressing the result as a percentage,24.2%percent of legal quarters is rejected.

Yes, the percentage is significantly high as the probability is more than 0.05 to reject the legal quarter.

Find the z score corresponding to the given value of area

b.

The vending machines are adjusted to accept all legal quarters except those who have weight in the top 2.5% and bottom 2.5%.

Let $x_1,x_2$ be the weights of quarters, such that $x_1$ is maximum weight of bottom 2.5% quarters and$x_2$ is the minimum weight of top 2.5% quarters.

Thus,

$$\begin{gathered} P\left(X<x_1\right)=P\left(Z<z_1\right) \\ =0.025 \\ P\left(X>x_2\right)=P\left(Z>z_2\right) \\ =0.025 \end{gathered}$$

For$x_1$, the cumulative area from the left is 0.025. In the statistical table for negative z score of Normal distribution, find 0.025; then the corresponding row value is -1.9 and column values is 0.06.This corresponds to the value of-1.96,which is the value of $z_1$.

For $x_2$, the cumulative area from the left is:

$$\begin{gathered} P\left(X>x_2\right)=1-P\left(X<x_2\right) \\ P\left(X<x_2\right)=0.975 \\ P\left(Z<z_2\right)=0.975 \end{gathered}$$

In the statistical table for positive z score of Normal distribution, find 0.975; then the corresponding row value is 1.9 and column values is 0.06.This corresponds to the value of 1.96, which is the value of $z_2$.

Find the extreme values

Thus, using the formula for z-scores:

$$\begin{gathered} x_1=\mu +z_1\times \sigma \\ =5.67+\left(-1.96\right)\times 0.06 \\ =5.5524 \end{gathered}$$

$$\begin{gathered} x_2=\mu +z_2\times \sigma \\ =5.67+1.96\times 0.06 \\ =5.7876 \end{gathered}$$

Therefore,the top 2.5% weights would weigh over5.7876 g, while the bottom 2.5% weigh below 5.5524g.

The limits for weights for quarters to be accepted are between 5.55g and 5.79 g.