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?

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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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 90, \sigma = 21[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{53 - 90}{21}[/tex]

[tex]Z = -1.76[/tex]

[tex]Z = -1.76[/tex] has a pvalue of 0.0392

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