contestada

Assume that adults have IQ scores that are normally distributed with a mean of 98.8 and a standard deviation 17.1. Find the first quartile Q1​, which is the IQ score separating the bottom​ 25% from the top​ 75%. (Hint: Draw a​ graph.)

Respuesta :

kudzordzifrancis

Answer:

The  IQ score separating the bottom​ 25% from the top​ 75%. is 87.3

Step-by-step explanation:

Given that adults have IQ scores that are normally distributed with a mean of 98.8 and a standard deviation 17.1.

To find the first quartile, we calculate X such that P(X<z)=0.25.

From the normal distribution table the The z-value that corresponds to an area of 0.25 is z=-0.675

We  use the formula:

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

We substitute to get:

[tex]-0.675=\frac{X-98.8}{17.1}[/tex]

This implies that:

[tex]17.1*-0.675=X-098.8[/tex]

[tex]-11.5425=X-098.8[/tex]

Solve for x to get:

[tex]x=98.8-11.5425[/tex]

X=87.2575

Using the normal distribution, it is found that the IQ score separating the bottom​ 25% from the top​ 75% is of 87.

------------------------

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula, which  in a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], is given by:

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

  • Each z-score has a p-value associated, which we find looking at the z-table, and this p-value represents the percentile of the measure X.

------------------------

  • Mean of 98.8 means that [tex]\mu = 98.8[/tex]
  • Standard deviation of 17.1 means that [tex]\sigma = 17.1[/tex];
  • The value of the interest is the 25th percentile, which 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 - 98.8}{17.1}[/tex]

[tex]X - 98.8 = -0.675(17.1)[/tex]

[tex]X = 87[/tex]

The IQ score is 87.

A similar problem is given at https://brainly.com/question/21628677