Each morning John eats some eggs. On any given morning, the number of eggs he eats is equally likelyto 1, 2, 3, 4, or 5 independent of what he has done in the past. Let X be the number of eggs that John eats in 10 days. Find the mean and the variance of X

Mean = 3

Variance = 2

Step-by-step explanation:

Since number of eggs he eats is equally likely to any 5 numbers, probability of each one will be (1/5) = 0.2

The probability mass function will therefore be

x 1 2 3 4 5

p 0.2 0.2 0.2 0.2 0.2

The mean is given as the expected value

And expected value is given as

E(X) = Σ xᵢpᵢ

where x = each possible sample space

p = probability of the sample space occurring.

E(X) = (1×0.2) + (2×0.2) + (3×0.2) + (4×0.2) + (5×0.2) = 3 eggs.

b) Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

Σx²p = (1² × 0.2) + (2² × 0.2) + (3² × 0.2) + (4² × 0.2) + (5² × 0.2) = 0.2 + 0.8 + 1.8 + 3.2 + 5 = 11

Var(X) = Σx²p − μ² = 11 - 3² = 11 - 9 = 2

the solution is h=10 because you add 9 to each side of the equation.