Question

question 4 of 5 > 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 σ = 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)

Explanation:

Step1: Find the z - score

We want the proportion of lids that are too small to be 0.01. Looking up the z - score in the standard normal distribution table (the area to the left of the z - score is 0.01), the z - score $z\approx - 2.3263$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $x = 3.95$ (the lower - limit of the acceptable diameter), $\sigma=0.02$, and $z=-2.3263$. We need to solve for $\mu$.
Rearranging the formula for $\mu$ gives $\mu=x - z\sigma$.

Step3: Calculate the mean

Substitute $x = 3.95$, $z=-2.3263$, and $\sigma = 0.02$ into the formula:
$\mu=3.95-(-2.3263)\times0.02=3.95 + 2.3263\times0.02=3.95+0.046526 = 3.9965$.

Answer:

$3.9965$