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If the distribution is really (5.43,0.54)

, what proportion of observations would be less than 5.79?
(Enter your answer rounded to four decimal places.)

User Jason Kao
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1 Answer

5 votes

Answer:

0.7486 = 74.86% observations would be less than 5.79

Explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be 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.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:


\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So


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


Z = (5.79 - 5.43)/(0.54)


Z = 0.67


Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

User Danivicario
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5.1k points