HW8-key

STAT 516 PRACTICE QUESTION SET 8 (WITH KEY) (1) Defective items on an assembly line occur independently with probability 0.05. A random sample of 100 items is taken. What is the probability that the first item sampled is not defective, given that at least 99 of the sampled items are NOT defective. Let A = { First is not defective } , B = { At least 99 are not defective } , then P ( A | B ) = P ( A, B ) P ( B ) = 0 . 95 · [ ( 99 98 ) 0 . 95 98 · 0 . 05 + 0 . 95 99 ] ( 100 99 ) (0 . 95) 99 · 0 . 05 + 0 . 95 100 = 99 · 0 . 05 + 0 . 95 5 + 0 . 95 = 5 . 9 5 . 95 ≈ 0 . 991597 . (2) A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the probability that fewer than 2 of the selected balls are red. Let A = { Fewer than 2 of the selected balls are red } . Then P ( A ) = 0 . 7 5 + 5 · (0 . 7) 4 · (0 . 3) = 0 . 52822 . (3) A small commuter plane has 30 seats. The probability that any particular passenger will not show up for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available. Let A = { 32 or 31 passengers show up } , then P ( A ) = 0 . 9 32 + 32 · 0 . 9 31 · 0 . 1 = 0 . 9 31 (0 . 9 + 3 . 2) = 0 . 15642 . 1

(4) The distribution of loss sizes has distribution function F ( x ) =      1 - ( 1000 x ) 1 . 2 , x > 1000 0 , otherwise Calculate the probability that at least 4 out of 6 losses exceed mean loss size. The mean is E ( X ) = Z ∞ 1000 ( 1000 x ) 1 . 2 dx + 1000 = - 5(1000) 1 . 2 x - 0 . 2 | ∞ 1000 + 1000 = 6000 . You can also calculate the mean by differentiating F ( x ) with respect to x and take integration. The probability for one loss to exceed the mean is 1 - F (6000) = 1 6 1 . 2 = 0 . 116471, hence the total probability p = 6 4 ( 1 6 1 . 2 ) 4 (1 - ( 1 6 1 . 2 )) 2 + 6 5 ( 1 6 1 . 2 ) 5 (1 - ( 1 6 1 . 2 )) 1 + ( 1 6 1 . 2 ) 6 = 0 . 002271 . (5) Auto claim amounts, in thousands, are modeled by a random variable with density function f ( x ) = xe - x for x ≥ 0. The company expects to pay 100 claims if there is no deductible. How many claims does the company expect to pay of the company decides to introduce a deductible of 1000? The number of claims is binomial with n = 100 and p = 1 - F (1), the probability that a claim is greater than 1000. We need the probability that a claim is greater than 1000. We have F ( x ) = Z x 0 f ( u ) du = Z x 0 ue - u du = - xe - x + 1 - e - x . Plugging in x = 1 for 1000 (since claim amounts are in thousands), F (1) = 1 - 2 e - 1 = 0 . 264241 . The expected number of claims if there is a deductible of 1000 is np = 100(1 - 0 . 2642) = 74. (6) An electronic system contains three cooling components that operate independently. The probability of each component’s failure is 0.05. The system will overheat if and only if at least two components fail. Calculate the probability that the system will overheat. A = { The system overheats } , then P ( A ) = (0 . 05) 3 + 3 2 (0 . 95)(0 . 05) 2 = 7 . 25 × 10 - 3 . 2

By the relation between negative binomial and geometric distributions, we have Var( Y ) = 5Var( X ) = 15 / 4 . (11) Let X have a Poisson distribution with λ = 1. What is the probability that X ≥ 2 given that X ≤ 4. By definition of conditional probability, P X ≥ 2 | X ≤ 4 = P [2 ≤ X ≤ 4] P [ X ≤ 4] = e - 1 (1 / 2 + 1 / 6 + 1 / 24) e - 1 (1 + 1 + 1 / 2 + 1 / 6 + 1 / 24) = 17 / 24 65 / 24 = 17 65 . (12) The number of traffic accidents per week in a small city has a Poisson distribution with mean equal to 3. What is the probability of exactly 2 accidents in 2 weeks? The number of traffic accidents in 2 weeks has a Poisson distribution with twice the weekly parameter, or λ = 6. So the probability is e - λ λ 2 / 2! = 18 e - 6 . (13) The number of power surges in an electric grid has a Poisson distribution with a mean of 1 power surge every 12 hours. What is the probability that there will be no more than 1 power surge in a 24-hour period? The Poisson parameter for a 24-hour period is twice the Poisson parameter for a 12-hour period, or equivalently λ = 2. The probability of 0 or 1 power surges in 24 hours is e - λ + λe - λ = e - 2 + 2 e - 2 = 3 e - 2 . (14) An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims. The number of claims filed has a Poisson distribution. Calculate the variance of the number of claims filed. Let N ∼ Poisson( λ ), we know P N [2] = 3 P N [4]. So, e - λ λ 2 2! = 3 e - λ λ 4 4! λ 2 2 = λ 4 8 λ 2 = 4 λ = 2 We obtain Var( N ) = λ = 2. 4

(15) Let X , Y and Z be independent Poisson random variables with E [ X ] = 3, E [ Y ] = 1 and E [ Z ] = 4. What is P [ X + Y + Z ≤ 1]? By the expectation formula of Poisson distribution, we readily know X ∼ Poisson(3), Y ∼ Poisson(1) and Z ∼ Poisson(4). We have S = X + Y + Z ∼ Poisson(8) and P [ S ≤ 1] = e - 8 + 8 e - 8 = 9 e - 8 . (16) Let X have a binomial distribution with parameters n and p , and let the conditional distribution of Y given X = x be Poisson with mean x . What is the variance of Y ? By the conditional variance formula, Var( Y ) = Var( E [ Y | X ]) + E [Var( Y | X )] = Var( X ) + E [ X ] = np (1 - p ) + np = np (2 - p ) . (17) The number of workplace injuries, N , occurring in a factory on any given day is Poisson distributed with mean Λ. Meanwhile, Λ is a random variable that is determined by the level of activity in the factory, and is uniformly distributed on the interval [0, 3]. Calculate Var( N ). By the conditional variance formula, Var( N ) = E[Var( N | Λ)] + Var(E[ N | Λ]) = E[Λ] + Var(Λ) = 3 / 2 + 3 2 / 12 = 2 . 25 . 5