From past record, the final exam marks out of 100 of a particular subject are we

Question

From past record, the final exam marks out of 100 of a particular subject are well approximated by a normal distribution with a mean of 60.

(a) If 5% of students achieve high distinction and high distinction mark is set at 85 and above, what is the standard deviation of the mark distribution?

(b) If less than 50 marks are considered as fail, what is the implied pass rate?

(c) If more than 55 marks to 65 marks are considered as “credit ”, what proportion of students would achieve “credit”? What mark a student must obtain to be in the top 10 percentile?

(d) In a class of 50 students, what proportion of student would achieve “credit”?

(e) Do you need to assume normality of the mark distribution for calculation of part (d)? Explain your answer. If the standard deviation of the of the population standard deviation as determined in (a) is not known, what would be the best estimator.

Explanation / Answer

as we know that z =(X-mean)/std deviation

and for highest 5% i.e at 95% ; z =1.64485

hence a) (85-60)/std deviation =1.64485

std deviation =15.20

b) for pass rate P(X>=50) =1-P(X<50) =1-P(Z<(50-60)/15.20) =1-P(Z<-0.6579) =1-0.2553 =0.7447

hence implied pass rate =74.47%

C) P(55<X<65) =P((55-60)/15.20<Z<(65-60)/15.20) =P(-0.329<Z<0.329) =0.6289 -0.3711 =0.2578

for top 10% ; z =1.28155

hence score X =mean+z*std deviation =60+15.20*1.28155 =79.48

d) In a class of 50 students proportion of student would achieve “credit” =0.2578

e)yes as per central limit theorum, as n goes large more then 40, it follows normal distribution

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.