Normal Distribution: Statistics Question

Question

Suppose birth weight is normally distributed with a mean of 7.5 pounds and a standard deviation of 1 pound. What is the probability that out of 5 randomly selected babies, exactly 4 of them are heavier than 8 pounds?

Answer 1

This is a normal distribution with the parameters μ = 7.5 pounds and σ = 1 pound. We need to use two things to solve this problem: use the binomial probability formula and the z-score of a normal distribution.

1.) Find the probability of one baby weighing more than 8 pounds.

z = (X - μ) / σ = (8 - 7.5) / 1 = 0.5/1 = 0.5

P(z > 0.5) = 1 - P(z ≤ 0.5) = 1 - 0.6915 = 0.3085

2.) Apply the binomial probability formula. We have a binomial distribution with the parameters of n = 5 babies and p = 0.3085 (success probability of babies weighing more than 8 pounds). x represents the numbers of babies selected from the set.

P(X = x) = C(n, x)*p^x (1 - p)^n-x = n! / [x!*(n - x)!]*p^x*(1 - p)^{n - x}

P(X = 4) = C(5, 4)*(0.3085)⁴ (1 - 0.3085)^{5 - 4} = 5! / [4!*(5 - 4)!]*(0.3085)⁴*(0.6915) = 0.0313

The probability that exactly 4 out of 5 babies are heavier than 8 pounds is about 0.0313.