The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.14 in. The width of a door is normally distributed with a mean of 23.87 in and a standard deviation of 0.08 in. Assume independence of the two widths. Find the probability that the width of the casing exceeds the width of the door by more than 0.25 in.

0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.14 in.

This means that

The width of a door is normally distributed with a mean of 23.87 in and a standard deviation of 0.08 in.

This means that .

Find the probability that the width of the casing exceeds the width of the door by more than 0.25 in?

This is P(C - D > 0.25).

Distribution C - D:

The mean is:

The standard deviation is:

Probability:

This probability is 1 subtracted by the pvalue of Z when X = 0.25. So

has a pvalue of 0.7704

1 - 0.7704 = 0.2296

0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.

its line 2 i just added the rest or else it wouldn't let me answer