Question

Suppose a normal distribution has a mean of 38 and a standard deviation of

Answer 1

Answer:

77.4% probability that a data value is between 36 and 41

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 question, we have that:

$\mu = 38, \sigma = 2$

What is the probability that a data value is between 36 and 41?

This is the pvalue of Z when X = 41 subtracted by the pvalue of Z when X = 36.

X = 41

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

$Z = \frac{41 - 38}{2}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.933

X = 36

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

$Z = \frac{36 - 38}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.933 - 0.159 = 0.774

77.4% probability that a data value is between 36 and 41