Scores on an examination are assumed to be normally distributed with mean 78 and

Question

Scores on an examination are assumed to be normally distributed with mean 78 and variance 36. a) What is the probability that a person taking the examination scores higher than 72? b) Suppose that students scoring in the top 10% of this distribution are to receive an A grade. What is the minimum score a student must achieve to earn an A grade? c) What must be the cutoff for passing the examination if the examiner wants only the top 28.1% of all scores to be passing? d) Approximately what proportion of students have scores 5 or more points above the score that cuts off the lowest 25%? e) If it is know that a student's score exceeds 72, what is the probability that his or her score exceeds 84?

Explanation / Answer

Q.3 Mean = 78 and variance = 36

so , standard devition = 6

(a) so NormPr( X>= 72; 78; 6) =1 - Pr( X< 72; 78; 6)

Z - value = (72 - 78)/6 = -1

respectve P - value for z= -1 => 0.1587

so NormPr( X>= 72; 78; 6) = 1 - 0.1587 = 0.8413

(b) In this question, we have to identify an X for which any value occuring above it have probability 0.1.

NormPr( x>=X; 78; 6) = 0.1

so NormPr( x>=X; 78; 6) = 1- NormPr( x< X; 78; 6)

NormPr( x< X; 78; 6) = 0.9

so from Z_ table, value of Z fr P - value 0.9 is 1.28

so 1.28 = ( X - 78)/6

X = 85. 68 marks

(c) so Here in this case NormsPr( x>=X; 78; 6) = 0.2810

=> NormsPr( x< X; 78; 6) = 1- 0.2810 = 0.7190

searching Z - table we will find Z - value = 0.58

so 0.58 = ( X - 78)/6

X = 78 + 0.58 * 6 = 81.48

(d) The cut off score of lowest 25% or say NormsPr( x <=X ; 78; 6) = 0.25

bychecking Z - value from Z - table for given probability is equal to - 0.675

so ( X - 78)/6 = - 0.675

X = 73.95

so 5 or more points above this score is equal to 78.95

so NormsPr( X> 78.95 ; 78; 6) = 1- NormsPr( X =< 78.95 ; 78; 6)

Z Value = (78.95 - 78)/6 = 0.1583

so respective p- value = 0.5630

so,  NormsPr( X> 78.95 ; 78; 6) = 1- 0.5630 = 0.437

(e) Pr( X >= 84/ X>= 72; 78; 6) = NormsPr(X >= 84; 78; 6)/ NOrmsPr(X >= 72; 78; 6)

we will calculate Z - value for both

so we can see that Z- values for X = 72 and X = 84 is +- 1

so, NormsPr(X >= 84; 78; 6)/ NOrmsPr(X >= 72; 78; 6) = ( 1- 0.8413)/ 0.8413 = 0.1886

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