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)
"Given \\; that, \u03bc=48, \u03c3=5, n=100, then,\\\\\ni)P(43<x<53) = P(\\frac{43-48}{5}< Z <\\frac{53-48}{5})\\\\\n=P(-1<Z<1)\\\\=2 P(0<Z<1)\\\\\n=2(0.3413)=0.6826\\\\\n\\therefore \\text{number of students}=0.6826 \\times 100\\\\\n=68.26 \\approx 69\\\\\nii)P(x>40) = P( Z >\\frac{40-48}{5})\\\\\n=P(Z>-1.6)\\\\=0.5+ P(0<Z<1.6)\\\\\n=0.5+0.4452=0.9452\\\\\n\\therefore \\text{number of students}=0.9452\\times 100\\\\\n=94.52\\approx 95\\\\\niii)P(x>50) = P( Z >\\frac{50-48}{5})\\\\\n=P(Z>0.4)\\\\=0.5- P(0<Z<0.4)\\\\\n=0.5-0.1554=0.3446\\\\\n\\therefore \\text{number of students}=0.3446\\times 100\\\\\n=34.46\\approx 35\\\\"
Comments
Leave a comment