Question #152010
Based on a sample of 450 AUS students, we found the average score is 165, with a standard deviation of 6.
a) At least how many percent of students scored between 150 to 180.
b) Assuming the histogram of GPA has a right skewed distribution, calculate the range of GPA that will cover at least 71% of the students?
c) Assuming the histogram of GPA follows the normal distribution, approximately how many students score between 147 to 159?
1
Expert's answer
2020-12-21T17:23:49-0500

Let X=X= the score of the student: XN(μ,σ2)X\sim N(\mu, \sigma^2)

Then Z=XμσN(0,1)Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)

Given μ=165,σ=6\mu=165, \sigma=6

a)


P(150<X<180)=P(X<180)P(X150)P(150<X<180)=P(X<180)-P(X\leq150)

=P(Z<1801656)P(Z1501656)=P(Z<\dfrac{180-165}{6})-P(Z\leq\dfrac{150-165}{6})

=P(Z<2.5)P(Z2.5)=P(Z<2.5)-P(Z\leq-2.5)

0.9937900.0062100.9876\approx0.993790-0.006210\approx0.9876

98.76%98.76\%


b)


P(Xx)=1P(Z<x1656)=0.71P(X\geq x)=1-P(Z<\dfrac{x-165}{6})=0.71

P(Z<x1656)=0.29P(Z<\dfrac{x-165}{6})=0.29

x16560.553385\dfrac{x-165}{6}\approx-0.553385

x=1656(0.553385)=162x=165-6(0.553385)=162

71% of the students will receive the score no less than 162162


c)


P(147<X<159)=P(X<159)P(X147)P(147<X<159)=P(X<159)-P(X\leq147)

=P(Z<1591656)P(Z1471656)=P(Z<\dfrac{159-165}{6})-P(Z\leq\dfrac{147-165}{6})

=P(Z<1)P(Z3)=P(Z<-1)-P(Z\leq-3)

0.1586550.0013500.1573\approx0.158655-0.001350\approx0.1573

15.73%15.73\%




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