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3. The adult men of the Dinaric Alps have the highest average height of all regions. The
distribution of height is approximately normal with a mean height of 6 ft 1 in (73 inches)
and standard deviation of 3 inches.
Find the 40th percentile of the height of Dinaric Alps distribution for men.
a.
b. What would be the minimum height of man in the Dinaric Alps that would place
him in the top 10% of all heights?
4. On a recent math test, the mean score was 75 and the standard deviation 5. Mike scored a
93.
What is the 25th percentile for the math scores (This is the same as Qi)?
a.
b. What is the 75th percentile for the math scores (This is the same as Q3)?

Respuesta :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • 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 [tex]\mu = 73[/tex].
  • The standard deviation is of 3 inches, hence [tex]\sigma = 3[/tex].

Item a:

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.253 = \frac{X - 73}{3}[/tex]

[tex]X - 73 = -0.253(3)[/tex]

[tex]X = 72.2[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 73}{3}[/tex]

[tex]X - 73 = 1.28(3)[/tex]

[tex]X = 76.84[/tex]

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 [tex]\mu = 75[/tex].
  • The standard deviation is of 5, hence [tex]\sigma = 5[/tex].

Item a:

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 75}{5}[/tex]

[tex]X - 75 = -0.675(5)[/tex]

[tex]X = 71.6[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.675 = \frac{X - 75}{5}[/tex]

[tex]X - 75 = 0.675(5)[/tex]

[tex]X = 78.4[/tex]

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 https://brainly.com/question/24663213

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