Question

b) In a examination, the mean score was 48 and standard deviation was 5 for a group

of 100 students. Assuming a normal distribution,

i) how many students scored between 43 and 53?           (2 marks)

ii) how many scored above 40?                                       (2 marks)

iii) how many scored above 50?                                      (2 marks)

Answer 1

$Given \; that, μ=48, σ=5, n=100, then,\\ i)P(43<x<53) = P(\frac{43-48}{5}< Z <\frac{53-48}{5})\\ =P(-1<Z<1)\\=2 P(0<Z<1)\\ =2(0.3413)=0.6826\\ \therefore \text{number of students}=0.6826 \times 100\\ =68.26 \approx 69\\ ii)P(x>40) = P( Z >\frac{40-48}{5})\\ =P(Z>-1.6)\\=0.5+ P(0<Z<1.6)\\ =0.5+0.4452=0.9452\\ \therefore \text{number of students}=0.9452\times 100\\ =94.52\approx 95\\ iii)P(x>50) = P( Z >\frac{50-48}{5})\\ =P(Z>0.4)\\=0.5- P(0<Z<0.4)\\ =0.5-0.1554=0.3446\\ \therefore \text{number of students}=0.3446\times 100\\ =34.46\approx 35\\$