Answer to Question #130759 in Statistics and Probability for tito

Question #130759

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)


1
Expert's answer
2020-08-26T17:51:23-0400

Given  that,μ=48,σ=5,n=100,then,i)P(43<x<53)=P(43485<Z<53485)=P(1<Z<1)=2P(0<Z<1)=2(0.3413)=0.6826number of students=0.6826×100=68.2669ii)P(x>40)=P(Z>40485)=P(Z>1.6)=0.5+P(0<Z<1.6)=0.5+0.4452=0.9452number of students=0.9452×100=94.5295iii)P(x>50)=P(Z>50485)=P(Z>0.4)=0.5P(0<Z<0.4)=0.50.1554=0.3446number of students=0.3446×100=34.4635Given \; 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\\


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