Question

Each morning, Hungry Harry eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4, 5, or 6, independently of what he has done in the past. Let X be the number of eggs that Harry eats in 10 days. Find the mean and variance of X.

Answer 1

Answer:

Mean = 35

Variance = 291.7

Step-by-step explanation:

Data provided in the question:

X : 1, 2, 3, 4, 5, 6

All the data are independent

Thus,

The mean for this case will be given as:

Mean, E[X] = $\frac{\textup{Sum of all the observations}}{\textup{Total number of observations}}$

or

 E[X] = $\frac{\textup{1+2+3+4+5+6}}{\textup{6}}$

or

E[X] = 3.5

For 10 days, Mean = 3.5 × 10 = 35

And,

variance = E[X²] - ( E[X] )²

Now, for this case of independent value,

E[X²] = $\frac{1^2+2^2+3^2+4^2+5^2+6^2}{\textup{6}}$

or

E[X²] = $\frac{1+4+9+16+25+36}{\textup{6}}$

or

E[X²] = $\frac{91}{\textup{6}}$

or

E[X²] = 15.167

Therefore,

variance = E[X²] - ( E[X] )²

or

variance = 15.167 - 3.5²

or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

= 291.7