Given that "\\mu=151,\\sigma=15,n=500"
Let X be the random variable denoting the weight of the students.
a) Then we have to find "P(120<X<155)"
Let Z be the standard normal variable, then
"Z=(X-\\mu)\/\\sigma"
When "X=120"
"Z=(120-151)\/15"
"=-2.07"
When "X=155"
"Z=(155-151)\/15"
"=0.27"
"P(120<X<155)=P(-2.07<Z<0.27)"
"=P(Z<0.27)-P(Z<-2.07)"
"=0.606-0.019=0.587"
Now the number of required student will be "(500*0.587)\\approx244"
b)
We have to find "P(X>18.5)"
Let Z be the standard normal variable, then
"Z=(X-\\mu)\/\\sigma"
When "X=18.5"
"Z=(18.5-151)\/15"
"=-8.83"
"P(X>18.83)=P(Z>-8.83)"
"=1-P(Z<-8.83)"
"\\approx1"
Now the number of required student will be "(500*1)=500"
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