A distribution of values is normal with a mean of 232.4 and a standard deviation of 92.2. Find the probability that a randomly selected value is between 66.4 and 241.6. P(66.4 < X < 241.6)

P(66.4 < X < 241.6) = 0.5039 = 50.39%.

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.

A distribution of values is normal with a mean of 232.4 and a standard deviation of 92.2.

This means that

Find the probability that a randomly selected value is between 66.4 and 241.6.

This is the p-value of Z when X = 241.6 subtracted by the p-value of Z when X = 66.4.

X = 241.6

has a p-value of 0.5398

X = 66.4

has a p-value of 0.0359

0.5398 - 0.0359 = 0.5039

P(66.4 < X < 241.6) = 0.5039 = 50.39%.

the answer is (-4,-4)

hope this ! : d