Question

If the mean of a normally distributed population is -10 with a standard deviation of 2, what is the likelihood of obtaining a value less than or equal to -7?

Answer 1

Answer:

93.32% probability of obtaining a value less than or equal to -7.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$Z = \frac{X - \mu}{\sigma}$

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 problem, we have that:

$\mu = -10, \sigma = 2$

What is the likelihood of obtaining a value less than or equal to -7?

This is the pvalue of Z when X = -7.

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{-7 -(-10)}{2}$

$Z = \frac{3}{2}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332.

So there is a 93.32% probability of obtaining a value less than or equal to -7.