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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?

User Dementrock
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Answer:

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

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 = (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 = (X - \mu)/(\sigma)


Z = (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

User Herrfz
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