Question

A normal population has a mean of 19 and a standard deviation of 5.

Answer 1

Answer:

a) $Z = 1.2$

b) 38.49% of the population is between 19 and 25.

c) 34.46% of the population is less than 17.

Step-by-step 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

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

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. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.

For this problem, we have that

A normal population has a mean of 19 and a standard deviation of 5, so $\mu = 19, \sigma = 5$.

(a) Compute the z value associated with 25

This is Z when $X = 25$

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

$Z = \frac{25 - 19}{5}$

$Z = 1.2$

(b) What proportion of the population is between 19 and 25?

This is the pvalue of Z when $X = 25$ subtracted by the pvalue of Z when $X = 19$.

X = 25 has $Z = 1.2$, that has a pvalue of 0.8849.

X = 19 has $Z = 0$, that has a pvalue of 0.5000.

So 0.8849-0.500 = 0.3849 = 38.49% of the population is between 19 and 25.

(c) What proportion of the population is less than 17?

This is the pvalue of Z when $X = 17$

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

$Z = \frac{17 - 19}{5}$

$Z = -0.40$

$Z = -0.40$ has a pvalue of 0.3446.

This means that 34.46% of the population is less than 17.