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erik [133]
3 years ago
12

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

r deciliter (1/10 of a liter) of blood. Suppose that after a 12-hour fast, the random variable x will have a distribution that is approximately normal with mean = 87 and standard deviation = 25. Note: After 50 years of age, both the mean and standard deviation tend to increase. For an adult (under 50) after a 12-hour fast, find the following probabilities. (Round your answers to four decimal places.)
(a) x is more than 60



(b) x is less than 110


(c) x is between 60 and 110


(d) x is greater than 125 (borderline diabetes starts at 125)
Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

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 <u>1 subtracted by the p-value of Z when X = 60</u>, 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 <u>p-value of Z when X = 110</u>, 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 <u>p-value of Z when X = 110 subtracted by the p-value of Z when X = 60</u>.

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 <u>1 subtracted by the p-value of Z when X = 125</u>, 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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