Question

A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 60 minutes and a standard deviation of 10 minutes. A random sample of eight service calls is taken. What is the probability that exactly three of them take more than 68.4 minutes? For calculation convenience, you may round probability results to the nearest thousandth (3 decimals).  

Answer 1

Here, μ = 60, σ = 10 and x = 68.4. We need to compute P(X >= 68.4). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (68.4 - 60)/10 = 0.84

Therefore,
P(X >= 68.4) = P(z <= (68.4 - 60)/10)
= P(z >= 0.84)
= 1 - 0.7995 = 0.2005

Here, n = 8, p = 0.2005, (1 - p) = 0.7995 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 8C3 * 0.2005^3 * 2^5
P(X = 3) = 0.147
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