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Suppose you are told that the grades of the nal exam of your class follows a normal distribution. The average score is 65 and the variance is 12. What is the the probability of a random classmate getting a score more than 80?

User BuschnicK
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1 Answer

4 votes

Answer:

0% probability of a random classmate getting a score more than 80

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
\mu and standard deviation(which is the square root of the variance)
\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 question, we have that:


\mu = 65, \sigma = √(12) = 3.4641

What is the the probability of a random classmate getting a score more than 80?

This is 1 subtracted by the pvalue of Z when X = 80. So


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


Z = (80 - 65)/(3.4641)


Z = 4.33


Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability of a random classmate getting a score more than 80

User Tyrese
by
7.3k points
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