Answer to Question #309040 in Statistics and Probability for paul

Question

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

We have

$\left( i \right) P\left( X<30 \right) =P\left( \frac{X-45}{14}<\frac{30-45}{14} \right) =P\left( Z<-1.07143 \right) =\\=\varPhi \left( -1.07143 \right) =0.1420\Rightarrow n=0.1420\cdot 100\approx 14\\\left( ii \right) P\left( 30\leqslant X\leqslant 70 \right) =P\left( \frac{30-45}{14}\leqslant \frac{X-45}{14}\leqslant \frac{70-45}{14} \right) =\\=P\left( -1.07143\leqslant Z\leqslant 1.78571 \right) =\varPhi \left( 1.78571 \right) -\varPhi \left( -1.07143 \right) =\\=0.96293-0.14200=0.82093\Rightarrow n=0.82093\cdot 100\approx 82\\\left( iii \right) P\left( 60\leqslant X\leqslant 80 \right) =P\left( \frac{60-45}{14}\leqslant \frac{X-45}{14}\leqslant \frac{80-45}{14} \right) =\\=P\left( 1.07143\leqslant Z\leqslant 2.5 \right) =\varPhi \left( 2.5 \right) -\varPhi \left( 1.07143 \right) =\\=0.99379-0.858012=0.135778\\\left( iv \right) P\left( X\geqslant 60 \right) =P\left( \frac{X-45}{14}\geqslant \frac{60-45}{14} \right) =\\=P\left( Z\geqslant 1.07143 \right) =1-\varPhi \left( 1.07143 \right) =\\=1-0.858012=0.141988\\\left( v \right) P\left( X\geqslant 40 \right) =P\left( \frac{X-45}{14}\geqslant \frac{40-45}{14} \right) =\\=P\left( Z\geqslant -0.357143 \right) =1-\varPhi \left( -0.357143 \right) =\\=1-0.360492=0.639508\\$