At some fast-food restaurants, customers who want a lid for their drinks get them from a large stack near the straws, napkins, and condiments. The lids are made with a small amount of flexibility so they can be stretched across the mouth of the cup and then snugly secured. When lids are too small or too large, customers can get very frustrated, especially if they end up spilling their drinks. At one particular restaurant, large drink cups require lids with a “diameter” of between $3.95$ and $4.05$ inches. The restaurant’s lid supplier claims that the diameter of its large lids follows a Normal distribution with a mean of $3.98$ inches and a standard deviation of $0.02$ inches. Assume that the supplier’s claim is true.

Put a lid on it!

a. What percent of large lids are too small to fit?

b. What percent of large lids are too big to fit?

c. Compare your answers to parts (a) and (b). Does it make sense for the lid manufacturer to try to make one of these values larger than the other? Why or why not?

Lid diameter,$x=3.95$

Mean,$\mu =3.98$

Standard deviation,$\sigma =0.02$

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

Calculate the $z-$score,

$z=\frac{x-\mu }{\sigma }=\frac{3.95-3.98}{0.02}=-1.50$

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

See the usual normal probability table for $P(z<-1.50)$and the row that starts with $-1.5$and the column that starts with$.00$

$$\begin{gathered} P(x<3.95)=P(z<-1.50) \\ =0.0668 \\ =6.68\% \end{gathered}$$

Thus,

Approximately $6.68$ percent of large lids will not fit.

Calculate the $Z-$score,

$z=\frac{x-\mu }{\sigma }=\frac{4.05-3.98}{0.02}=3.50$

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

$3.50$do not exist in the table, but $3.49$exist.

See the typical normal probability table for $P(z<3.50)$and the row that starts with $3.5$and the column that starts with$.00$

$$\begin{gathered} P(x>4.05)=P(z>3.50) \\ =1-P(z<3.50)\approx 1-P(z<3.49) \\ =1-0.9998 \\ =0.0002 \\ =0.02\% \end{gathered}$$

Thus

Approximately $0.02\%$ of the large lids are too big to fit.

According to Part (a) result,

Approximately $6.68$ percent of large lids will not fit.

According to Part $\left(b\right)$result,

Approximately$0.02\%$ of the large lids are too big to fit.

Note that

In comparison to the lids that will be very large, there are a lot more little lids.

Hence,

It's pointless to try to make one of these values larger than the other because the lids will not fit in either scenario (they won't fit when tooled tiny, and they won't fit when tooled large).