Question

question 2 of 10
at fast - food restaurants, the lids for drink cups 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 frustrated, especially if they end up spilling their drinks. at one restaurant, large drink cups require lids with a diameter of between 3.95 and 4.05 inches. the restaurants lid supplier claims that the diameter of the large lids follows a normal distribution with mean 3.98 inches and standard deviation 0.02 inch. assume that the suppliers claim is true. the supplier is considering two changes to reduce the percent of its large - cup lids that are too small to 1%: (1) adjusting the mean diameter of its lids, or (2) altering the production process to decrease the standard deviation of the lid diameters.
(a) if the standard deviation remains at $sigma = 0.02$ inch, at what value should the supplier set the mean diameter of its large - cup lids so that only 1% are too small to fit?
mean =
inches
(round to 4 decimal places.)
(b) what effect will the change in part (a) have on the percent of lids that are too large?
by increasing the mean from 3.98 inches to the value i entered in part (a), the percentage of lids that are too large will

Explanation:

Step1: Use z - score formula

We know that for a normal distribution $X\sim N(\mu,\sigma^{2})$, the z - score is given by $z=\frac{x - \mu}{\sigma}$. We want to find the mean $\mu$ when the area to the left of $x = 3.95$ is $0.01$. Looking up the z - score in the standard normal distribution table (the $z$ value corresponding to an area of $0.01$ to the left), we get $z=- 2.3263$.

Step2: Rearrange z - score formula to solve for $\mu$

We have $z=\frac{x - \mu}{\sigma}$, and we want to solve for $\mu$. Rearranging gives $\mu=x - z\sigma$. Substituting $x = 3.95$, $z=-2.3263$ and $\sigma = 0.02$ into the formula.
\[

$$\begin{align*} \mu&=3.95-(-2.3263)\times0.02\\ &=3.95 + 2.3263\times0.02\\ &=3.95+0.046526\\ &=3.9965 \end{align*}$$

\]

Step3: Analyze the effect on large - sized lids

The normal distribution is symmetric. When we increase the mean, the area to the right of the upper - bound (4.05 inches) will increase. This is because the entire distribution shifts to the right. So the percentage of lids that are too large will increase.

Answer:

(a) $3.9965$
(b) increase