Question

In an intelligence test administered on 1000 children, the average was 60 and the standard deviation was 20. Assuming that the marks obtained by the children follow a normal distributed, find the number of children who have scored (i) over 90 marks, (ii) below 40 marks and (iii) between 50 and 80 marks. Given that P(0 < Z < 1.5) = 0.4332; P(Z < 1) = 0.8413; P(Z < 0.5) = 0.6915. 

Answer 1

(i)

$$P(X>90)=1-P(Z\le\dfrac{90-60}{20})$$

$$=1-P(Z\le1.5)=1-(0.5+0.4332)=0.0668$$

$0.0668(1000)=67$

(ii)

$$P(X<40)=P(Z<\dfrac{40-60}{20})=P(Z<-1)$$

$$=1-P(Z<1)=1-0.8413=0.1587$$

$0.1587(1000)=159$

(iii)

$$P(50<X<80)=P(Z<\dfrac{80-60}{20})$$

$$-P(Z\le\dfrac{50-60}{20})=P(Z<1)-P(Z\le-0.5)$$

$$=0.8413-(1-0.6915)=0.5328$$

$0.5328(1000)=533$