Question

Drying times for newly painted microwave oven cabinets are normally distributed with a mean of 2.5 minutes and a standard deviation of 0.25 minutes. After painting, each cabinet is mated with its electronic modules and mechanical components. The production manager must decide how much time to allow after painting before these other components are installed. If the time is too short, the paint will smudge and the unit will have to be refinished. If the time is too long, production efficiency will suffer. A consultant has concluded that the time delay should be just enough to allow 99.8% of the cabinets to dry completely, with just 0.2% ending up being smudged and sent back for refinishing. Given this information, for what time setting should the production manager set the automatic timer that pauses the production line while each cabinet dries?

Answer 1

To determine the appropriate time for the production manager's automatic timer, use the standard normal distribution to find the drying time corresponding to the upper tail area of 0.2%.

Step-by-step explanation:

To determine the appropriate time setting for the production manager's automatic timer, we need to find the value that corresponds to the upper tail area of 0.2%. This is because the production manager wants to allow enough time for 99.8% of the cabinets to dry completely. Since the drying times are normally distributed with a mean of 2.5 minutes and a standard deviation of 0.25 minutes, we can use the standard normal distribution to find the corresponding z-score.

Using a standard normal distribution table or a calculator, we find that the z-score corresponding to an upper tail area of 0.2% is approximately 2.77. To find the corresponding drying time, we can use the formula:

z = (x - mean) / standard deviation

Solving for x:

x = z * standard deviation + mean

Substituting the values:

x = 2.77 * 0.25 + 2.5

x ≈ 3.1925

Therefore, the production manager should set the automatic timer to pause the production line for approximately 3.19 minutes while each cabinet dries.

Answer 2

Answer: 3.22

Step-by-step explanation:

Given that;

mean = 2.5 min

standard deviation = 0.25 min

now to get the value of X required, we say

z = (x - u) / a

where z is the distance from the mean measured in the standard deviation units, x is the value we are interested in, u is the mean distribution, a is the standard deviation of the distribution.

the time delay should be just enough to allow 99.8% of the cabinets to dry completely = 99.8/100 = 0.9980

first we determine an appropriate z value.

Using the standardized normal tables,

value of z for approximately 0.9980 is 2.88

so using our initial equation z = (x - u) / a

we substitute the value

z = (x - u) / a

2.88 = ( x - 2.5) / 0.25

2.88 * 0.25 = x - 2.5

0.72 = x - 2.5

x = 0.72 + 2.5

x = 3.22