Mathematics, 19.03.2021 05:00, TH3L0N3W0LF

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

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answered: wehack6380

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 33, \sigma = 4$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{37 - 33}{4}$

Z = 1

Z = 1 has a pvalue of 0.8413

X = 29

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29 - 33}{4}$

Z = -1

Z = -1 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 33}{4}$

Z = 3

Z = 3 has a pvalue of 0.9987

X = 33

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33 - 33}{4}$

Z = 0

Z = 0 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29 - 33}{4}$

Z = -1

Z = -1 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{37 - 33}{4}$

Z = 1

Z = 1 has a pvalue of 0.8413

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

answered: Guest

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

answered: Guest

570, the answer is 570, hope it

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A normal distribution has a mean of 33 and a standard deviation of 4. Find the probability that a ra...