Question

Put a lid on it! Refer to Exercise $56$. The supplier is considering two changes to reduce the percent of its large-cup lids that are too small to less than $1\%$. One strategy is to adjust the mean diameter of its lids. Another option is to alter the production process, thereby decreasing the standard deviation of the lid diameters.

(a) If the standard deviation remains at $\sigma =0.02$inches, at what value should the supplier set the mean diameter of its large-cup lids to ensure that less than $1\%$are too small to fit? Show your method.

(b) If the mean diameter stays at $\mu =3.98$inches, what value of the standard deviation will result in less than $1\%$of lids that are too small to fit? Show your method.

(c) Which of the two options in (a) and (b) do you think is preferable? Justify your answer. (Be sure to consider the effect of these changes on the percent of lids that are too large to fit.)

Short answer

a) The mean lid diameter should be $4.00$inches to ensure that less than $1\%$of the lids are too small to fit

b) The proportion of observation less than $3.5$is approximately $1$. so the proportion of observations greater than $3.5$is approximately $0$.

c) we prefer reducing the standard deviation. this will not increase the number of lids that are too small

Step-by-step solution

Part (a) Step-1 Given Information

Standard deviation$\sigma =0.02$

The mean diameter of its large-cup lids ensures that less than $1\%$are too small to fit.

Part (a) Step-2 Explanation

For example, if the standard deviation for the large-cup lids is $0.02$inches, then the mean diameter of the lids needs to be set so that less than $1\%$of them are too small to fit.

Our goal is to find a mean which leaves less than $1\%$of lids with a diameter smaller than$3.95.$Calculate the $z-$value for which$1\%$of observations fall below.

The $z$value of $0.01(1\%)$is determined by using the standard normal table.

Therefore, it is$-2.33$.

In order to find the $z-$value that corresponds to a diameter of$3.95$, we solve:

localid="1649933016724" $$\begin{gathered} -2.33=\frac{3.95-\mu }{0.02} \\ \mu =4.00inches \end{gathered}$$

To ensure less than $1\%$of the lids are too small, the mean diameter of the lids should be $4.00$inches.

Part (b) Step-1: Given Information

The mean diameter stays at $\mu =3.98$inches

The standard deviation will result in less than $1\%$of lids that are too small to fit.

Part (b) Step-2: Explanation

Using $3.98$as the mean diameter, then we should determine the value of the mean diameter that the supplier should set for the lids of large cups to ensure that fewer than $1\%$of them are too small to fit.

In order to ensure that less than $1\%$of lids have a diameter that is less than $3.95$, we need to calculate the proper standard deviation.

The first step is to find the $z-$value for which$1\%$of the observations are below.

The $z$value for $0.01(1\%)$, as determined by the standard normal table, is therefore $0.01$.

That is$z=-2.33$

$$\begin{gathered} -2.33=\frac{3.95-3.98}{\sigma } \\ \sigma =0.013 \end{gathered}$$

In this case, the proportion of observations greater than $3.5$is approximately $0$, so the proportion of observations less than $3.5$is approximately $1$. Therefore, it is extremely rare for the lid to be too big.

Part (c) Step-1: Given Information

Given in the question that

a) The mean lid diameter should be $4.00$inches to ensure that less than $1\%$of the lids are too small to fit

b) The proportion of observation less than $3.5$is approximately 1. so the proportion of observations greater than $3.5$is approximately $0$we have to find out that Which of the two options in (a) and (b) do you think is preferable

Part (c) Step-2: Explanation

The standard deviation should be reduced. Therefore, there will be fewer lids that are too big, and fewer that are too small. The number of lids that are too small will decrease if we shift the mean to the right, but we will increase the number of lids that are too big.