Let "X" be the weight of a student, and random variable "Z" has standard normal distribution "Z" ~"N(0,1)" .
We assume that "X" has normal distribution, then "X" ~"N(40,5^2)", and
"\\frac{X-40}{5}" ~"Z".
The probability that student weight ">50" is
"p=P(X>50)=P(\\frac{X-40}{5}>\\frac{50-40}{5})=\\\\\nP(Z>2)=1-P(Z\\leq2)=\\\\1-0.9772=0.0228."
Let "k" be the number of students weighing ">50", then "k" has a binomial distribution
with the probability of success "p=0.0228" and parameter "n=5".
a)
"p(k=5)=C_n^kp^k(1-p)^{n-k}=\\\\\nC_5^5*0.0228^5*0.9772^0=0.0228^5\\approx 6*10^{-9}."
b)
"p(k=3)=C_n^kp^k(1-p)^{n-k}=\\\\\nC_5^3*0.0228^3*0.9772^2=\\\\\n\\frac{5*4*3}{1*2*3}*0.0228^3*0.9772^2\\approx 10^{-4}."
Answer: a) "6*10^{-9}" , b) "10^{-4}."
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