Question

Score on the standardized test are approximately normally distributed with a mean of 480 and a standard deviation of 90.

Six students are chosen at random. What is the probability that exactly one of them scores more than 600?

Answer 1

Here, μ = 480, σ = 90 and x = 600. We need to compute P(X >= 600). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (600 - 480)/90 = 1.33

Therefore,
P(X >= 600) = P(z <= (600 - 480)/90)
= P(z >= 1.33)
= 1 - 0.9082
= 0.0918

Now, n = 6, p = 0.0918, (1 - p) = 0.9082 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 1)
P(X = 1) = 6C1 * 0.0918^1 * 0.9082^5
P(X = 1) = 0.3403