Question

An insurance company has found that 8% of its claims are for damages resulting from burglaries. What is the probability that a random sample of 20 claims will contain 5 or fewer that are for burglary damages?

a. What is the probability that a sample of 20 claims will contain more than 5 claims for damages resulting from burglaries?

b. What is the expected number of claims resulting from burglaries?

Answer 1

X ~ Binomial (n,p)

Where n = 20 , p = 0.08

P(X) = ⁿC_x p^x (1-p)^n-x

So,

P( X <= 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)

= ²⁰C₀ 0.08⁰ 0.92²⁰ + ²⁰C₁ 0.08¹ 0.92¹⁹ + ²⁰C₂ 0.08² 0.92¹⁸ + ²⁰C₃ 0.08³ 0.92¹⁷ + ²⁰C₄ 0.08⁴ 0.92¹⁶ + ²⁰C₅ 0.08⁵ 0.92¹⁵

= 0.9962

a)

P( X > 5) = 1 - P (X <= 5)

= 1 - [ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P( X = 5) ]

= 1 - [ ²⁰C₀ 0.08⁰ 0.92²⁰ + ²⁰C₁ 0.08¹ 0.92¹⁹ + ²⁰C₂ 0.08² 0.92¹⁸ + ²⁰C₃ 0.08³ 0.92¹⁷ + ²⁰C₄ 0.08⁴ 0.92¹⁶

+ ²⁰C₅ 0.08⁵ 0.92¹⁵]

= 1 - 0.9962

= 0.0038

b)

E(X) = n * p

= 20 * 0.08

= 1.6