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The distribution of scores on the SAT is approximately normal with a mean of mu = 500 and a standard deviation of sigma = 100. For the population of students who have taken the SAT, a.What proportion have SAT scores greater than 700? b.What proportion have SAT scores greater than 550? c.What is the minimum SAT score needed to be in the highest 10% of the population? d.If the state college only accepts students from the top 60% of the SAT distribution, what is the minimum SAT score needed to be accepted ?

Respuesta :

jennerfranco

Answer:

a. 2.28%

b. 30.85%

c. 628.16

d. 474.67

Step-by-step explanation:

For a given value x, the related z-score is computed as z = (x-500)/100.

a. The z-score related to 700 is (700-500)/100 = 2, and P(Z > 2) = 0.0228 (2.28%)

b. The z-score related to 550 is (550-500)/100 = 0.5, and P(Z > 0.5) = 0.3085 (30.85%)

c. We are looking for a value b such that P(Z > b) = 0.1, i.e., b is the 90th quantile of the standard normal distribution, so, b = 1.281552. Therefore, P((X-500)/100 > 1.281552) = 0.1, equivalently  P(X > 500 + 100(1.281552)) = 0.1 and the minimun SAT score needed to be in the highest 10% of the population is 628.1552

d. We are looking for a value c such that P(Z > c) = 0.6, i.e., c is the 40th quantile of the standard normal distribution, so, c = -0.2533471. Therefore, P((X-500)/100 > -0.2533471) = 0.6, equivalently P(X > 500 + 100(-0.2533471)), and the minimun SAT score needed to be accepted is 474.6653

Using the normal distribution, it is found that:

a) 0.0228 = 2.28% of students have SAT scores greater than 700.

b) 0.3085 = 30.85% of students have SAT scores greater than 550.

c) The minimum SAT score needed to be in the highest 10% of the population is 628.

d) The minimum SAT score needed to be accepted is 475.

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.

In this problem:

  • The mean is [tex]\mu = 500[/tex]
  • The standard deviation is [tex]\sigma = 100[/tex].

Item a:

This proportion is 1 subtracted by the p-value of Z when X = 700, thus:

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

[tex]Z = \frac{700 - 500}{100}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

0.0228 = 2.28% of students have SAT scores greater than 700.

Item b:

This proportion is 1 subtracted by the p-value of Z when X = 550, thus:

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

[tex]Z = \frac{550 - 500}{100}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a p-value of 0.6915.

1 - 0.6915 = 0.3085.

0.3085 = 30.85% of students have SAT scores greater than 550.

Item c:

This is the 100 - 10 = 90th percentile, which 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 - 500}{100}[/tex]

[tex]X - 500 = 1.28(100)[/tex]

[tex]X = 628[/tex]

The minimum SAT score needed to be in the highest 10% of the population is 628.

Item d:

This is the 100 - 60 = 40th percentile, which is X when Z has a p-value of 0.4, so X when Z = -0.25.

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

[tex]-0.25 = \frac{X - 500}{100}[/tex]

[tex]X - 500 = -0.25(100)[/tex]

[tex]X = 475[/tex]

The minimum SAT score needed to be accepted is 475.

A similar problem is given at https://brainly.com/question/24663213