Now let X be the random variable indicating the number of students present.
Now it is given that the probability of a student being present is 30% so,
Now X follows binomial distribution with parameters "\\left(n,\\:p\\right)=\\left(450,\\:0.3\\right)"
Now as n approaches infinity and p is finite so we use binomial approximation to normal distribution.
"np=450\\times 0.3=135;"
"\\sqrt{\\left(np\\left(1-p\\right)\\right)}=\\sqrt{\\left(135\\times 0.7\\right)}"
We are required to find,
"\\:P\\left(X>150\\right)=P\\left(X>150.5\\right)\\:\\left[for\\:discrete\\:to\\:continuous\\:approximation\\right]"
"=P\\frac{\\left(Z>\\left(150.5-135\\right)\\right)}{\\sqrt{\\left(135\\times 0.7\\right)}}"
"=P\\left(Z>1.594\\right)"
"=1-P\\left(Z<1.59\\right)"
"=1-\\left(0.5+0.441\\right)"
"=0.059"
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