Answer to Question #302990 in Statistics and Probability for anu

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)

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

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

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

$0.5328(1000)=533$