Answer to Question #161250 in Quantitative Methods for Sunny

Question #161250

Assume that the test scores from a college admissions test are normally distributed, with a

mean of 450 and a standard deviation of 100.

a. What percentage of the people taking the test score between 400 and 500?

b. Suppose someone receives a score of 630. What percentage of the people taking the

test score better? What percentage score worse?

c. If a particular university will not admit anyone scoring below 480, what percentage of

the persons taking the test would be acceptable to the university?


1
Expert's answer
2021-02-24T06:43:36-0500

We have:

μ=450,σ=100\mu=450,\sigma=100


a.

z1=x1μσ=400450100=0.5z_1=\frac{x_1-\mu}{\sigma}=\frac{400-450}{100}=-0.5

z2=x2μσ=500450100=0.5z_2=\frac{x_2-\mu}{\sigma}=\frac{500-450}{100}=0.5

P(400<x<500)=P(0.5<z<0.5)=2P(z<0.5)=20.3085=61.70%P(400<x<500)=P(-0.5<z<0.5)=2P(z<-0.5)=2\cdot0.3085=61.70\%


b.

z=xμσ=630450100=1.8z=\frac{x-\mu}{\sigma}=\frac{630-450}{100}=1.8

P(x<630)=P(z<1.8)=0.9641=96.41%P(x<630)=P(z<1.8)=0.9641=96.41\%

P(x>630)=P(z>1.8)=1P(z<1.8)=10.9641=0.0359=3.59%P(x>630)=P(z>1.8)=1-P(z<1.8)=1-0.9641=0.0359=3.59\%


c.

z=xμσ=480450100=0.3z=\frac{x-\mu}{\sigma}=\frac{480-450}{100}=0.3

P(x>480)=P(z>0.3)=1P(z<0.3)=10.6179=0.3821=38.21%P(x>480)=P(z>0.3)=1-P(z<0.3)=1-0.6179=0.3821=38.21\%


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