A curious student in a large economics course is interested in calculating the percentage of his classmates who scored lower than he did on the GMAT; he scored 490. He knows that GMAT scores are normally distributed and that the average score is approximately 540. He also knows that 95% of his classmates scored between 400 and 680. Based on this information, calculate the percentage of his classmates who scored lower than he did?

Respuesta :

Answer:  23.89%

Step-by-step explanation:

The empirical rule says that the 95% of the data falls in between two standard deviations of the mean.

Given : GMAT scores are normally distributed and the the average score is approximately [tex]\mu=540[/tex].

Also, 95% of his classmates scored between 400 and 680.

Then, by empirical rule , 95% of data falls in between [tex]mu\pm 2\sigma[/tex]

i.e. [tex]540- 2\sigma=400[/tex]    (1)

[tex]540+2\sigma=680[/tex]       (2)

Subtracting (1) from (2), we get

[tex]4\sigma=680-400=280\\\\\Rightarrow\ \sigma=\dfrac{280}{4}=70[/tex]

Let x be the random variable to represent the scores of every student.

Statistic z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 490, we have

[tex]z=\dfrac{490-540}{70}\approx-0.71[/tex]

The p-value = [tex]P(z<-0.71)=0.2388521\approx23.89\%[/tex]

Hence, 23.89% of his classmates who scored lower than he did .

23.89% of his classmates scored lower than he did.

Given

GMAT scores are normally distributed and the average score is approximate.

Also, 95% of his classmates scored between 400 and 680.

What is the empirical rule?

The empirical rule says that, in a normal data set, virtually every piece of data will fall within three standard deviations.

By empirical rule, 95% of data fall in between;

[tex]\rm= m\pm2\sigma[/tex]

Therefore,

[tex]\rm 540-2\sigma=400\\\\540+2\sigma=680[/tex]

Subtracting (1) from (2), we get

[tex]4\sigma =280\\\\\sigma = \dfrac{280}{4}\\\\\sigma=70[/tex]

Let x be the random variable to represent the scores of every student.

Then,

The value of the z-score is;

[tex]\rm z-score=\dfrac{x-\mu}{\sigma}\\\\ z-score=\dfrac{490-540}{70}\\\\z-score=\dfrac{-50}{70}\\\\z-score=-0.71[/tex]

The value of p is 23.89.

Hence, 23.89% of his classmates scored lower than he did.

To know more about the Empirical formula click the link given below.

https://brainly.com/question/11588623