Answer to Question #309358 in Statistics and Probability for isaac

Question

QUESTION THREE

In a Competitive examination of 5000 students, the marks of the examinees in statistics were found to be distributed normally with mean 45 and standard deviations 14.

Determine the number of examinees whose marks, out of 100 were;

(i) Less than 30.                                                         2MKS

(ii) Between 30 and 70.                                              2MKS

(iii) Between 60 and 80.                                          2MKS

(iv) More than 60.                                                 2MKS

(v) More than 40                              2MKS

 

 

Answer 1

$\mu=45,\sigma=14,Z=(x-μ)/σ$

i. $P(x<30)=P(Z<(30-45)/14)=P(Z<-1.07)$

From z distribution table

$P(Z<-1.07)=0.1423$

therefore number of students $=5000 * 0.1423 = 712$

ii. $P(30<x<70)=P((30-45)/14<Z<(70-45)/14)$

$=P(-1.07<Z<1.79)=P(Z<1.79)-P(Z<-1.07)$

$=0.9633-0.1423=0.821$

Number of students $=5000*0.821=4105$

iii. $P(60<x<80)=P((60-45)/14<Z<(80-45)/14)$

$=P(1.07<Z<2.5)=P(Z<2.5)-P(Z<1.07)$

$=0.9946-0.8577=0.1369$

Number of students $=5000*0.1369=685$

iv. $P(x>60)=P(Z>(60-45)/14)=P(Z>1.07)=1-P(Z<1.07)$

$=1-0.8577=0.1423$

Number of students $=5000*0.1423=712$

v. $P(x>40)=P(Z>(40-45)/14)=P(Z>-0.36)=1-P(Z<-0.36)$

$=1-0.3594=0.6406$

Number of students $=5000*0.6406=3203$