Question

The mean score of 2000 students appearing for an examination is 34.4 and the standard deviation is 16.6. How many candidates may be expected to obtain marks
i) between 30 and 60
ii) less than 32
iii)greater than 36
iv)between 36 and 39.

Answer 1

Assume the normality of distribution

Let $X=$ the number of candidates: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=34.4, \sigma=16.6,n=2000$

i)

$$P(30\leq X\leq60)=P(X\leq60)-P(X<30)=$$

$$=P(Z\leq\dfrac{60-34.4}{16.6})-P(Z<\dfrac{30-34.4}{16.6})$$

$$\approx P(Z\leq1.54219)-P(Z<-0.26506)$$

$$\approx0.938486-0.395482\approx0.543004$$

$2000\cdot0.543004=1086$

There are 1086 candidates between 30 and 60.

ii)

$$P(X<32)=P(Z<\dfrac{32-34.4}{16.6})$$

$$\approx (Z<-0.14458)\approx0.442521$$

$2000\cdot0.442521=885$

There are 885 candidates less than 32.

iii)

$$P(X>36)=1-P(X\leq36)$$

$$=1-P(Z\leq\dfrac{36-34.4}{16.6})$$

$$\approx1-(Z<0.09639)\approx1-0.538393\approx0.461607$$

$2000\cdot0.461607=923$

There are 923 candidates greater than 36.

iv)

$$P(36\leq X\leq39)=P(X\leq39)-P(X<36)=$$

$$=P(Z\leq\dfrac{39-34.4}{16.6})-P(Z<\dfrac{36-34.4}{16.6})$$

$$\approx P(Z\leq0.27711)-P(Z<0.09639)$$

$$\approx0.609152-0.538393\approx0.070759$$

$2000\cdot0.070759=142$

There are 142 candidates between 36 and 39.