Chapter 2.2, Problem 58E

(a)

To determine

To find: the value of the mean diameter of the large cups.

Answer to Problem 58E

The mean lid diameter should be 4.00 inches

Explanation of Solution

Given:

The standard deviation remains at $\sigma =0.02$ inches

Calculation:

If the standard deviation is 0.2 inches, then it is supposed to find the value of mean diameter that a supplier must set for the large-cup lids to ensure that less than 1% of the lids are too small to fit.

Solve for the appropriate mean so that less than 1% of lids have diameter less than 3.95. First, find the z -value for which 1% of observations are lower. Using standard normal table, find the z value corresponding to $0.01\left(1\%\right)$ . That is, $z=-2.33$ .

Since this z -value should correspond to a diameter of 3.95,

  $\begin{array}{l}-2.33=\frac{3.95-\mu }{0.02}\\ \mu =4.00\text{inches}\end{array}$

So the mean lid diameter should be 4.00 inches to ensure that less than 1% of the lids are too small to fit.

Conclusion:

Therefore, the mean lid diameter should be 4.00 inches

(b)

To determine

To find: the value of the standard deviation of the lids which are very small to fit.

Answer to Problem 58E

It is extremely rare for a lid to be too big.

Explanation of Solution

Given:

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

Calculation:

If the mean diameter stays at 3.98, then find the value of mean diameter that a supplier must set for the large-cup lids to ensure that less than 1% of the lids are too small to fit.

Solve for the appropriate standard deviation so that less than 1% of lids have diameter less than 3.95. First, find the z -value for which 1% of observations are lower. Using standard normal table, find the z value corresponding to $0.01\left(1\%\right)$ . That is, $z=-2.33$ .

  $\begin{array}{l}-2.33=\frac{3.95-3.98}{\sigma }\\ \sigma =0.013\text{inches}\end{array}$

Hence, the proportion of observations less than 3.5 is approximately 1, so the proportion of observations greater than 3.5 is approximately 0. In other words, it is extremely rare for a lid to be too big.

Conclusion:

Therefore, it is extremely rare for a lid to be too big.

(c)

To determine

To find: whether option a or b is preferable.

Answer to Problem 58E

Part (b) is preferred.

Explanation of Solution

Given:

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

Calculation:

Reducing the standard deviation is preferred. This will not increase the number of lids that are too big, but will reduce the number of lids that are too small. If the mean is shifted to the right, reduce the number of lids that are too small, but increase the number of lids that are too big.

Conclusion:

Therefore, part (b) is preferred.