Question

3. The adult men of the Dinaric Alps have the highest average height of all regions. The

Answer 1

Using the normal distribution, it is found that:

• 3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

• 3 - b) The minimum height of man in the Dinaric Alps that would place  him in the top 10% of all heights is of 76.84 inches.

• 4 - a) The 25th percentile for the math scores was of 71.6 inches.

• 4 - b) The 75th percentile for the math scores was of 78.4 inches.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

Question 3:

• The mean is of 73 inches, hence $\mu = 73$.

• The standard deviation is of 3 inches, hence $\sigma = 3$.

Item a:

The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.

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

$-0.253 = \frac{X - 73}{3}$

$X - 73 = -0.253(3)$

$X = 72.2$

The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

Item b:

The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28.

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

$1.28 = \frac{X - 73}{3}$

$X - 73 = 1.28(3)$

$X = 76.84$

The minimum height of man in the Dinaric Alps that would place  him in the top 10% of all heights is of 76.84 inches.

Question 4:

• The mean score is of 75, hence $\mu = 75$.

• The standard deviation is of 5, hence $\sigma = 5$.

Item a:

The 25th percentile is X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 75}{5}$

$X - 75 = -0.675(5)$

$X = 71.6$

The 25th percentile for the math scores was of 71.6 inches.

Item b:

The 75th percentile is X when Z has a p-value of 0.25, so X when Z = 0.675.

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

$0.675 = \frac{X - 75}{5}$

$X - 75 = 0.675(5)$

$X = 78.4$

The 75th percentile for the math scores was of 78.4 inches.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213