Question

A laboratory equipment's supplier gets 45% of the electronic weighing scale supplies from country Z. Five electronic weighing scale are selected at random. Determine the mean and standard deviation for the number of electronic weighing supplies by country Z.

Answer 1

Explanation:

Let's denote X as the number of electronic weighing supplies from country Z out of the five selected scales. Since the supplier gets 45% of the supplies from country Z, we can conclude that X follows a binomial distribution with parameters n = 5 (the number of trials) and p = 0.45 (the probability of success, i.e., getting a scale from country Z).

a). Mean (μ):

The mean of a binomial distribution is given by the formula μ = n * p. Therefore, the mean number of electronic weighing supplies from country Z is:

μ = 5 * 0.45 = 2.25

b). Standard Deviation (σ):

The standard deviation of a binomial distribution is given by the formula σ = (n * p * (1 - p)). Therefore, the standard deviation for the number of electronic weighing supplies from country Z is:

σ = √(5 * 0.45 * (1 - 0.45))

σ ≈ √(2.475)

σ ≈ 1.57