The Student Room Group

s2 - question 2

The marks of 500 candidates in an exam are normally distributed with mean of 45 marks and a standard deviation of 20 marks.

a) given that pass mark is 41, estimate the number of candidates who passed the exam.

I got 0.5793 as the probability but how do you estimate the number of candidates with this?

b) If 5% of candidates obtain a distinction with x marks or more, estimate the value of x.

I got 77.9. Is that correct?

c) Estimate the interquartile range of the distribution.

I am stuck on this.

Rep awarded. :wink:
confused?
The marks of 500 candidates in an exam are normally distributed with mean of 45 marks and a standard deviation of 20 marks.

a) given that pass mark is 41, estimate the number of candidates who passed the exam.

I got 0.5793 as the probability but how do you estimate the number of candidates with this?



Let S be marks scored.

a) People who got less than 41 marks clearly failed.
Thus we can use 1-P(S<41) to give a proportion of the 500 candidates who passed.

Standardise S=41.

z = 41-45 / 20 = -4/20 = -0.2

Phi (-0.2) = 1-Phi(0.2) = 0.4207

Hense 0.5793 or 57.93 % of candidates passed.
ie 500 * 0.5793 = 289.65 candidates and since people are most definately discrete (!) this is 289 candidates.
confused?


b) If 5% of candidates obtain a distinction with x marks or more, estimate the value of x.

I got 77.9. Is that correct?



Phi^-1 (0.95) = 1.6449 = z

1.6449 = (x - 45)/20

20*1.6449 + 45 = x = 77.898 = 77.9 (1dp)
confused?


c) Estimate the interquartile range of the distribution.

I am stuck on this.

Rep awarded. :wink:


IQR = UQ - LQ

LQ = S: P(s<=S) = 0.25
UQ = S: P(s<=S) = 0.75

Phi^-1 (.25) = -0.6745
Phi^-1(.75) = 0.6745

For LQ:

-0.6745 = (x - 45)/20
=> x = 20 * -0.6745 + 45
= x = 31.51

For UQ:

0.6745 = (x - 45) / 20
=> x = 20 * 0.6745 + 45
=> x = 58.49

IQR = UQ - LQ = 58.49 - 31.51 = 26.98 marks.