A normal distribution has a mean of 33 and a standard deviation of 4. Find the probability that a randomly selected x-value from the distribution is in the given interval. a. between 29 and 37 b. between 33 and 45 c. at least 29 d. at most 37

a. 0.6826 = 68.26% probability that a randomly selected x-value from the distribution is between 29 and 37.

b. 0.4987 = 49.87% probability that a randomly selected x-value from the distribution is between 33 and 45.

c. 0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at least 29.

d. 0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at most 37.

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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

A normal distribution has a mean of 33 and a standard deviation of 4.

This means that

a. between 29 and 37

This is the pvalue of Z when X = 37 subtracted by the pvalue of Z when X = 29. So

X = 37

has a pvalue of 0.8413

X = 29

has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability that a randomly selected x-value from the distribution is between 29 and 37.

b. between 33 and 45

pvalue of Z when X = 45 subtracted by the pvalue of Z when X = 33. So

X = 45

has a pvalue of 0.9987

X = 33

has a pvalue of 0.5

0.9987 - 0.5 = 0.4987

0.4987 = 49.87% probability that a randomly selected x-value from the distribution is between 33 and 45.

c. at least 29

This is 1 subtracted by the pvalue of Z when X = 29. So

has a pvalue of 0.1587

1 - 0.1587 = 0.8413

0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at least 29.

d. at most 37

This is the pvalue of Z when X = 37. So

has a pvalue of 0.8413

0.8413 = 84.13% probability that a randomly selected x-value from the distribution is at most 37.

answer: y = -12x + 108

step-by-step explanation: do (y = m x+b) formula.

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