Question

Let X be the yearly claim (in dollars) of a randomly chosen policyholder of MAICO, an auto insurance company. It is known that E(X) = 750 and Var(X) = 62,500.

a) Show that P(250 ≤ X ≤ 1250) ≥ 3/4 [Hint: Chebyshev's inequality]
b) State the Central Limit Theorem for a sequence of independent, indentically distributed random variables X1, X2, ..., each having mean μ and variance σ 2.
c) Let Ẍ be the average claim of 10,000 randomly chosen MAICO policyholders. Show that P(745 ≤ Ẍ ≤ 755) ≃ 0.9544 [Note: You may ignore the continuity correction]
d) Prove the Central Limit Theorem exactly as you have stated it in (b)

Answer 1

The answers are as explained below:

Explanation:

E(X) = 750 and Var(X) = 62,500.

a. By chebychev’s inequality,

P (│x-μ│≥ k) ≤ σ²/k²

P (│x-750│≥ k) ≤ 62500/k²

Or P (│x-750│≤ R) ≥ 1- 62500/k²

Now, P (250 ≤ X ≤ 1250)

= P (250-750 ≤ X-750 ≤ 1250- 750)

= P (-500 ≤ X-750≤ 500)

Hence, P (│x-750│≤ 500) ≥ (1- 62500)/500²

Therefore, P (│x-750│≤ 3/4

b. Central Limit theorem (CLT)

Let X1, X2, X3,…, Xn be a sequence of independently and identically distributed (11a)

Random variables with mean µ and variance σ², and let X =∑_(i=0)^n▒Xi, then as n increases, the distribution of X, Fx(x) converges to the normal distribution N (n µ, n σ²).

c. N= 10000, µ=750, σ²=62500, σ= 250

Using the central limit theorem,

Zn= (ẍ- µ)/ (σ/√n)

P(745 ≤ Ẍ ≤ 755)= P((745-750)/(250/ √1000) ≤ Zn ≤ ((755-750)/ (250/ √1000))

= P (-2 ≤ Zn ≤ 2 )

=P (Zn≤ 2) - P (Zn≤ -2)

= 0.9773 - 0.0228

≃ 0.9545

d. Recall, ẍ =∑_(i=1)^n▒Xi/n

then, E (ẍ)= E{ ∑_(i=1)^n▒Xi/n}

∑_(i=1)^n▒Xi/n {E (Xi)}= ∑_(i=1)^n▒µ/n

(nµ)/ n= µ

E { ẍ }= µ

Also, Var (ẍ)= Var {∑_(i=1)^n▒Xi/n}

∑_(i=1)^n▒〖Var (Xi)〗 / n²

∑_(i=1)^n▒σ²/ n² = (n σ²)/n²

= σ²/n

Var (ẍ) = σ²/n

Thus, ẍ ≃ N (µ, σ²/n ) (Proved)