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

Marks: 6M

Year: Dec 2015

Given:

Mean = 2.5

S.D = 3.5

We have S.N.V

$$Z = \frac{X-m}{\sigma } = \frac{X-2.5}{3.5}$$

1) Now to find P(2 $\mathrm{\le}$ x $\mathrm{\le}$4.5)

In terms of z by using S.N.V

When x =3 , z = 2-2.5/3.5 = -0.14

And When x =4 , z = 4.5-2.5/3.5 = -0.57

2) $\mathrm{\therefore }$ P(2 $\mathrm{\le}$ x $\mathrm{\le}$4.5) = P(-0.14 $\mathrm{\le}$ z $\mathrm{\le}$ 0.57)

$\mathrm{\therefore }$ Area between (z = -0.14 and z=0.57) = Area between (z = 0 and z=0.14) + Area between (z = 0 and z=0.57)

= 0.0557 + 0.2157 =0.2714

3) To find:

P(-1.5 $\mathrm{\le}$ x $\mathrm{\le}$5.5)

In terms of z by using S.N.V

When x =-1.5 , z = -1.5-2.5/3.5 = -1.14

And When x =5.5 , z = 5.5-2.5/3.5 = -0.8

$\mathrm{\therefore }$ P(-1.5 $\mathrm{\le}$ x $\mathrm{\le}$5.5) = P(-1.14 $\mathrm{\le}$ z $\mathrm{\le}$ 0.8)

$\mathrm{\therefore }$ Area between (z = -1.14 and z=0.8) = Area between (z = 0 and z=1.14) + Area between (z = 0 and z=0.8)

= 0.3729 + 0.2881 =0.6610