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A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly

Home/Math/A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly

Question

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly selected score will be greater than 63.7

93.32% probability that a randomly selected score will be greater than 63.7.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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:

What is the probability that a randomly selected score will be greater than 63.7.

This is 1 subtracted by the pvalue of Z when X = 63.7. So

has a pvalue of 0.0668

1 – 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.

## Answers ( )

Answer:93.32% probability that a randomly selected score will be greater than 63.7.

Step-by-step explanation:Problems of normally distributed samples are solved using the z-score formula.In a set with mean and standard deviation , the zscore of a measure X is given by:

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:What is the probability that a randomly selected score will be greater than 63.7.This is 1 subtracted by the pvalue of Z when X = 63.7. So

has a pvalue of 0.0668

1 – 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.