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Mariulka [41]
3 years ago
13

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

per 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 μ = 90 and standard deviation of σ = 21 What is the probability that, for an adult after a 12-hour fast, x is less than 53?
Mathematics
1 answer:
omeli [17]3 years ago
7 0

Answer:

3.92% probability that, for an adult after a 12-hour fast, x is less than 53

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 90, \sigma = 21

What is the probability that, for an adult after a 12-hour fast, x is less than 53?

This is the pvalue of Z when X = 53.

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

Z = \frac{53 - 90}{21}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392

3.92% probability that, for an adult after a 12-hour fast, x is less than 53

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We substitute our given values into the expression above.

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Learn more about compound interest here: brainly.com/question/27128740

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