Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: May 2014

Mean(m) = 70

Standard deviation ($\sigma$) = 5

N = 1000

Let x denote marks obtained by a student :

1) P(between 60 and 75 marks) = P(60 $\lt X\lt75)$

$\mathrm{\therefore }$ We have $$z = \frac{x-m}{\sigma }$$ $$\mathrm{\therefore } z_1=\ \frac{60-70}{5}=\ -2$$ And $$z_2=\ \frac{75-70}{5}=1$$

$\mathrm{\therefore }$ Probability in terms of z will be $$P(-2 \lt z \lt 1)$$

$$\mathrm{\therefore } P(-2 \lt z \lt 1) = Area \ between \ z=0 \ to z=1 \ + \ Area \ between \ z=0 \ to \ z=-2 = 0.3413 + 0.4773 = 0.8186$$

$\mathrm{\therefore }$ Number of students scoring between 60 and 75 marks: $$N * P(-2 \lt z \lt 1) = 1000* 0.8186 = 819 students.$$

2) P(more than 75 ) = $$P(x \gt 75)$$

We have $$z = \frac{x-m}{\sigma }=\ \frac{75-70}{5}=1$$

Probablity in terms of z

$$\mathrm{\therefore }P(Z \gt 1) = 0.5 - Area \ between \ (z=0 to z=1) \ = 0.5 - 0.3413 = 0.1587$$

$\mathrm{\therefore }$ Number of students scoring more than 75 marks: $$N * P(X= x) = 1000 * 0.1587 = 159$$

Ans:

Number of students scoring between 60 to 75 is 819

Number of students scoring above 75 is 159