An egg supplier to a restaurant send a daily shipment of 500 eggs. Previous record show that 4% of these are eggs arrive to the restaurant damaged. X equals the number of eggs at arrived damaged on a randomly selected day. Find the mean and standard deviation of eggs

Mean = 20

Standard deviation ≈ 4.38

Step-by-step explanation:

X is a random variable with a binomial distribution with p = 4% = 0.04 and n = 500, where p stands for probability of success and n stands for number of trials.

1. The mean of binomial distribution is given as np, therefore the mean would be

500 x 0.04 = 20

2. The standard deviation of a binomial distribution is given as √[np(1 – p)], therefore we have:

√[500 x 0.04(1 – 0.04)] = √[20(0.96)] = √(19.2) ≈ 4.38

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