Given that μ=151,σ=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−μ)/σ
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)≈244
b)
We have to find P(X>18.5)
Let Z be the standard normal variable, then
Z=(X−μ)/σ
When X=18.5
Z=(18.5−151)/15
=−8.83
P(X>18.83)=P(Z>−8.83)
=1−P(Z<−8.83)
≈1
Now the number of required student will be (500∗1)=500
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