Question

A normal distribution has a mean of 10 and a standard

deviation of 1. What is the probability of selecting a
number that is at most 9?

Answer 1

15.87% probability of selecting a number that is at most 9

Explanation:

When the distribution is normal, we use 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, we have that:

`\mu = 10, \sigma = 1`

What is the probability of selecting a number that is at most 9?

This is the pvalue of Z when X = 9. So

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

`Z = (9 - 10)/(1)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

15.87% probability of selecting a number that is at most 9