Question

Prof. Weiner is interested in analyzing the distribution of the variable X=’grading in Microeconomics for Policy Analysis’. He collected administrative information from the School from 1990 to 2010 and based on this data he concludes X follows a normal distribution with mean 75 and variance 30.

2.1 What is the probability that a typical student from the current cohort enrolled in Microeconomics for Policy Analysis receives a grade above 80?

2.2 Between what values will the grades of 95% of all students in the current cohort fall?

2.3 Find the value that represents the 25^th and 75^th percentile of this distribution. Then, find the interquartile range. Can a student who got 45 be considered a potential outlier?  

Answer 1

a)

µ = 75     

σ² = 30

  

P ( X >80) = P( (X-µ)/σ ≥ (80-75)/√30)

= P(Z >0.91) = P( Z <-0.913) = 0.1807(answer)

2.2)

µ = 39.2      

σ =√30 = 1.9

proportion=0.9500

proportion left 0.0500is equally distributed both left and right side of normal curve

z value at0.025= ±1.960    (excel formula =NORMSINV(0.05/ 2 ) )

z = ( x - µ ) / σ

so, X = z σ + µ =

X1 =-1.960*1.9+39.2=35.4761

X2 =1.960*1.9+39.2=42.9239

2.3)

P(X≤x) =0.2500

  

Z value at 0.25=-0.6745(excel formula =NORMSINV(0.25) )

z=(x-µ)/σ

so, X=zσ+µ=-0.674*5.477225575+75

X=71.3057

so, 25th percentile = 71.31 (answer)   

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P(X≤x) =0.7500

  

Z value at 0.75=0.6745(excel formula =NORMSINV(0.75) )

z=(x-µ)/σ

so, X=zσ+µ=0.674*5.477225575+75

X=78.6943

75th percentile = 78.69 (answer)   

interquartile range = 75th percentile - 25th percentile = 78.69 - 71.31 = 7.39

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P( X ≤ 45) = P( (X-µ)/σ ≤ (45-75) /5.48)=P(Z ≤-5.48)

since, z score is greater than ±2 ,

so, yes, 45 be considered a potential outlier