What are the solutions to |5x+2|=8

A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose pe

soldier1979 [14.2K]

Using the normal distribution, it is found that:

a) 0.8599 = 85.99% probability that x is more than 60.

b) 0.1788 = 17.88% probability that x is less than 110.

c) 0.6811 = 68.11% probability that x is between 60 and 110.

d) 0.0643 = 6.43% probability that x is greater than 125.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

In this problem:

The mean is of 87, thus $\mu = 87$ .
The standard deviation is of 25, thus $\sigma = 25$ .

Item a:

This probability is 1 subtracted by the p-value of Z when X = 60, thus:

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

$Z = \frac{60 - 87}{25}$

$Z = -1.08$

$Z = -1.08$ has a p-value of 0.1401.

1 - 0.1401 = 0.8599

0.8599 = 85.99% probability that x is more than 60.

Item b:

This probability is the p-value of Z when X = 110, thus:

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

$Z = \frac{110 - 87}{25}$

$Z = 0.92$

$Z = 0.92$ has a p-value of 0.8212.

1 - 0.8212 = 0.1788.

0.1788 = 17.88% probability that x is less than 110.

Item c:

This probability is the p-value of Z when X = 110 subtracted by the p-value of Z when X = 60.

From the previous two items, 0.8212 - 0.1401 = 0.6811.

0.6811 = 68.11% probability that x is between 60 and 110.

Item d:

This probability is 1 subtracted by the p-value of Z when X = 125, thus:

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

$Z = \frac{125 - 87}{25}$

$Z = 1.52$

$Z = 1.52$ has a p-value of 0.9357.

1 - 0.9357 = 0.0643.

0.0643 = 6.43% probability that x is greater than 125.

A similar problem is given at brainly.com/question/24863330

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Answer:

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