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21. A gym has 2500 members. The ages of members have a normal distribution with mean of 47 years and standard deviation of 12
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Answer #1

Solution to 21)

Given μ = 47 and σ = 12

So,

μ - 2σ = 47 - 2(12) = 23

μ +2σ = 47 + 2(12) = 71

And two-sigma includes 95 percent, so the number of members aging between [23,71] is

= 2500*0.95 = 2375

Solution to 22)

X = μ + σZ = 47 + 12(2.2) = 73.4 years

Solution to 23)

Let p be the probability that person has blood Group O

n = 20 is sample size

Let the random variable X denotes the number of persons having blood Group O

So X ~ Binomial (20, 0.45)

So, P(X = x) = (nCx)p​​​​​​x(1-p)n-x

To find probability that atmost 6 person have blood Group O, so we need to find P(X ≤ 6)

P(X ≤ 6) = 0.1299337886

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