of securing total marks i) 180 or above ii)90 or below.

Solution:

Let $x_1 ,x_2 ,x_3 $ denotes the marks obtained in three subjects. The $x_1 ,x_2 ,x_3 $ are normal distribution with mean $51 ,53,46$ and variance $15^2 ,12^2,16^2$

Assuming the variates to be independent $y=x_1+x_2+x_3$ is distribbuted normally with mean $m= 51+53+46 =150$ and

$\sigma ^2=15^2+12^2+16^2$

$\sigma ^2=225+144+256=625$

$\sigma =\sqrt {625}=25$

$S.N.V Z= \frac{y-m}{\sigma}$

$Z= \frac{y-150}{25}$-------(1)

1) 180 or above ,

When y=180 eqn(1) becomes

$Z =\frac {180-150}{25}$

$Z=\frac{30}{25}$

$Z=1.2$

$P(y\ge 180)=P(Z \ge 1.2)$

=Area to the right of $Z=1.2$

=$0.5$ - (Area between $Z=0 \ to \ Z=1.2$)

=$0.5-03849$

$P(y\ge 180)=0.1151$

ii) 90 or below when y=90, eqn (1) becomes

$Z=\frac{90-150}{25}$

$Z=-2.4$

$P(y\le 90 )=P(Z\le -2.4)$

=Area to the left of $Z=-2.4$

=$0.5$ - Area from $Z=0 \ to \ Z=2.4$

=$0.5-0.4918$

$P(y\le 90 )=0.0082$