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Find the percent of all values in a normal distribution for which z ≤ 1.00, to the nearest tenth of percent.

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Answer:

84.1% of all values in a normal distribution have z ≤ 1.00.

Explanation:

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.

In this question:

The percent of all values in a normal distribution for which z ≤ 1.00.

This is the pvalue of Z = 1.

Z = 1 has a pvalue of 0.8413.

Converting to percentage, to the nearest tenth.

84.1% of all values in a normal distribution have z ≤ 1.00.

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