Question

Wooden poles produced for electricity networks in rural areas have lengths that are normally distributed. 2% of the poles are rejected because they are considered too short, and 5% are rejected because they are too long.a. Find the mean and standard deviation of these poles if the acceptable range is between

6.3 m and 7.5 m.b. In a randomly selected sample of 20 poles, find the probability of finding 2 rejected poles.

Answer 1

a. Let X be the length of a wooden pole produced for electricity networks in rural areas. The probability of a pole being rejected because it is too short is 0.02, and the probability of a pole being rejected because it is too long is 0.05. The acceptable range is between 6.3 m and 7.5 m. We can find the mean and standard deviation of X as follows:

First, we need to find the z-score corresponding to the lower bound of the acceptable range:

z1 = (6.3 - μ) / σ

Similarly, we need to find the z-score corresponding to the upper bound of the acceptable range:

z2 = (7.5 - μ) / σ

Using a standard normal table, we can find the values of z1 and z2 that correspond to the probabilities of 0.02 and 0.95, respectively:

z1 = -2.05

z2 = 1.64

Solving for μ and σ, we get:

z1 = (6.3 - μ) / σ => μ = 6.3 + 2.05σ

z2 = (7.5 - μ) / σ => σ = (7.5 - 6.3) / 1.64 = 0.74

Therefore, the mean and standard deviation of X are:

μ = 6.3 + 2.05(0.74) = 7.6 m

σ = 0.74 m

b. Let Y be the number of rejected poles in a sample of 20 poles. Y follows a binomial distribution with parameters n = 20 and p = 0.02 + 0.05 = 0.07. The probability of finding 2 rejected poles in a sample of 20 poles is:

P(Y = 2) = (20 choose 2) * 0.07^2 * 0.93^18 = 0.242

The probability of finding 2 rejected poles in a sample of 20 poles is 0.242.