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. 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? If less than 50 marks are considered as fall, what is the implied pass rate? 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? In a class of 50 students, what proportion of student would achieve "credit"? 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

Solution

Back-up Theory

If a random variable X N(µ, 2), then P(X or t) = P[Z or {(t - µ)/}] where Z = {(X - µ)/} is the Standard Normal Variate whose probability can be directly read off the Standard Normal Probability Table.

Now, to work out the solution,  

Let X = score in the exam. We are given X N(60, 2), where 2 is not known.

Part (a)

5% students get distinction (i.e., X 85) => P(X 85) = 0.05 or P[Z {(85 - 60)/}] = 0.05 or

P[Z (25/)] = 0.05....... (1) From Standard Normal Probability Table, P[Z 1.64] = 0.05...... (2)

(1) and (2) => (25/) = 1.64 or = 25/1.64 = 15.24 ANSWER

Part (b)

Pass rate = pass % = P(X 50) = P[Z {(50 - 60)/}] = P[Z {(- 10)/15.24}] = P[Z - 0.66]

= 1 - 0.2578 [From Standard Normal Probability Table}  

= 0.7422 or 74.22% ANSWER

Part (c) first part

'Credit' rate = P(55 X 65) = P[{(55 - 60)/} Z {(65 - 60)/}] = P[(- 5/15.24) Z (5/15.24)]

= P[- 0.33 Z 0.33] = (1 - 0.3707) - 0.3707 = 0.2586 [From Standard Normal Probability Table}

= 25.86% ANSWER

Part (c) second part

t is Top 10 percentile => P(X t) = 0.1 or P[Z {(t - 60)/15.24}] = 0.1 or {(t - 60)/15.24} = 1.28 or

t = 79.51 ANSWER

Part (d)

Proportion remains the same whether the number of students is 100 or 50. So, answer is 25.86% as obtained in first part of (c) above. [However, if the question were 'number of students getting 'credit', answer would be 50 x 0.2586 = 13]

Part (e)

As explained under Back-up Theory' all these calculations are bsed on Normal Distribution. Hence assumption of Normality is a must. And population standard deviation can be estimated by the sample standard deviation, where the divisor is (n - 1) and not n

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