0 x = 0.30 x = 1.10 x = 3.05 x = 4.15 x = x = 12. f(x) =

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1 . A mail-order computer business has si telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table p( Calculate the probability of each of the following events. a. {At most 3 lines are in use} P [X 3] P [X 0] P [X ] P [X 2] P [X 3] b. {Fewer than 3 lines are in use} P [X < 3] P [X 0] P [X ] P [X 2] c. {At least 3 lines are in use} P [X 3] P [X < 3] d. {Between 2 and 5 lines, inclusive, are in use} P [2 X 5] P [X 2] P [X 3] P [X 4] P [X 5] e. {Between 2 and 4 lines, inclusive, are not in use} P [X 2]P [X 3]P [X 4] f. {At least 4 lines are not in use} P [X 2] P [X 0] P [X ] P [X 2] An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X the number of months between successive payments. The cdf of X is as follows: 0 <.30 < < < < 4 F (.45 4 < < < < 2 2 a. What is the pmf of X? f(

2 b. Using just the cdf, compute P (3 < X < 6 and P (4 < X. P (3 < X < 6 P [X 3, 4, 6] P (4 < X P [X 6, 2] An appliance dealer sells three different models of upright freezers having 3.5, 5.9, and 9. cubic feet of storage space, respectively. Let X the amount of storage space purchased by the net customer to buy a freezer. Suppose that X has probability mass function p( a. Compute E(X, E(X 2, and V (X. E(X 6.38, E(X , and V (X b. If the price of a freezer having capacity X cubic feet is 25X 8.5, what is the epected price paid by the net customer to buy a freezer? E[25X 8.5] (25E[X] , c. What is the variance of the price 25X 8.5 paid by the net customer? V [25X 8.5] [(25 2 ]V [X] 2496 d. Suppose that while the rated capacity of a freezer is X, the actual capacity is h(x X.0X 2. What is the epected actual capacity of the freezer purchased by the net customer? E[h(X] E[X.0X 2 ] E[X] (.0E[X 2 ] A chemical supply company currently has in stock 00 Kg of a certain chemical, which it sells to customers 5 Kg lots. Let X the number of lots ordered by a randomly chosen customer, and suppose that X has pmf p( Compute E(X and V (X. Then compute the epected number of Kg s left after the net customer s order is shipped, and the variance of the number of Kg s left. 2

3 E[X] ( 0.2 (2 0.4 (3 0.3 ( V [X] E[X 2 ] (E[X] 2 ( 0.2(4 0.4(9 0.3( ( The required r.v. is Y 00 5X, therefore, E[Y ] 00 5 E[X] V [Y ] V [00 5 V [X]] (5 2 V [X] When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let X the number of defective boards in a random sample of size n 25, so X BIN(25, (a Determine P (X 2 (b Determine P (X 5 (c Determine P ( X 4 (d What is the probability that none of the 25 boards are defective? (e Calculate the epected value and standard deviation of X. (a Determine P [X 2]. ( 0 0 P [X 2] ( ( 0 ( 0 (0.5 [ 0.5] 0 ( ( !(0 0! [ ]!(0! [ ] 2!(0 2! [ ] ( ( (b Determine P [X 5]. 3

4 P [X 2] 0 5 ( 0 (0.5 [ 0.5] 0 5!(0 5! [ ] 6!(0 6! [ ] 7!(0 7! [ ] 8!(0 8! [ ] 9!(0 9! [ ] (0 0! [ ] ( (c Determine P [ X 4]. Required probability: ( Note that we could also have arrived at the same answer, without computing binomial coefficients, by observing that, P [ X 4] P [X 0] [ P [X 5]] (d What is the probability that none of the 0 boards are defective? This is the probability P [X 0] (e Calculate the epected value and standard deviation of X. We know that any Binomial r.v. can be defined as the sum of independent Bernoulli trials, i.e., if X BIN(n, p, then 4

5 X n i X i, where each X i BIN(, p, (or Bernoulli with parameter p - a r.v taking value with probability p and 0 with pr. p. Therefore E[X] E[ X i ] i E[X i ] np i V [X] V [ X i ] i V [X i ] i p( p np( p i Standard deviation is the square-root of variance, and hence equals Show that E(X np when X is a binomial random variable. [HINT: First epress E(X as a sum with lower limit. Then factor out np, let y so that the sum is from y 0 to y n, and show that the sum equals.] E[X] 0 ( n p ( p n n!!(n! p ( p n 0 n! (!(n! p ( p n n (n! n y!(n y! py ( p n y y0 np(p p n np 5

6 7. Customers at a gas station select either regular (A, premium (B, or diesel fuel (C. Assume that successive customers make independent choices, with P (A 0.3, P (B 0.2, and P (C 0.5. (a Among the net 00 customers, what are the mean and variance of the number who select regular fuel? Eplain your reasoning. The no. who select regular do so with P (A 0.3, therefore who do not select regular do so with P (A P (B P (C 0.7. Define X i if i-th customer chooses regular, 0 otherwise. This is therefore a Bernoulli r.v. with parameter, p 0.3. Therefore the total no. of customers among the 00 customers who choose regular, is a Binomial r.v. X i. Therefore the mean and variance are and (b wer part (a for the number among the 00 who select a nondiesel fuel. 8. An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maimum of si reservations for a trip, and a passenger must have a reservation. From previous records, 20% of all those making reservations do not appear for the trip. wer the following questions, assuming independence wherever appropriate. (a If si reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip? At least one individual with a reservation cannot be accommodated on the trip is equivalent to 5 or 6 individuals turning up. This is a Binomial r.v., as can be seen by defining X i if an individual turns up, and taking the total no. turning up as Y i X i. We now have to find, P [Y 5] P [Y 6]. ( ( ( ( (b If si reservations are made, what is the epected number of available places when the limousine departs? The available no. of places Z ma[4 Y, 0]. Therefore, P [Z k] P [Y 4 k], k,..., 4, and P [Z 0] P [Y 4] P [Y 5] P [Y 6]. The epected value is therefore,.p [Y 3] 2.P [Y 2] 3.P [Y ] 4.P [Y 0].P [Y 3] 2.P [Y 2] 3.P [Y ] 4.P [Y 0] 6

7 (c Suppose the probability distribution of the number of reservations made is given in the accompanying table. Number of reservations Probability Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X. Define the r.v. R to be the no. of reservations. First consider the case, X 0, this can happen even if R 3, 4, 5, 6. Therefore P [X 0] P [X 0, R 3]P [X 0, R 4]P [X 0, R 5]P [X 0, R 6] 3 ( ( ( ( ( ( ( ( P [X ] P [X, R 3]P [X, R 4]P [X, R 5]P [X, R 6] 3 ( ( ( ( ( ( ( ( P [X 2] P [X 2, R 3]P [X 2, R 4]P [X 2, R 5]P [X 2, R 6] P [X 3] P [X 3, R 3]P [X 3, R 4]P [X 3, R 5]P [X 3, R 6] P [X 4] P [X 0] P [X ] P [X 2] P [X 3] The article Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants (Water Research, 984, pp suggests the uniform distribution on the interval (7.5, 20 as a model for depth (cm of the bioturbation layer in sediment in a certain region. a. What are the mean and variance of depth? b. What is the cdf of depth? 7

8 c. What is the probability that observed depth is at most 0? Between 0 and 5? d. What is the probability that the observed depth is within SD of the mean value? Within 2 SD s? 0. Let X have the Pareto pdf given by { f(; k, θ a. If k >, compute E(X. kθ k k θ 0 < θ b. What can you say about E(X if k? c. If k > 2, show that V (X kθ 2 (k 2 (k 2 d. If k 2, what can you say about V (X? e. What conditions on k are necessary to ensure that E(X n is finite?. If bearing diameter is normally distributed, what is the probability that the diameter of a randomly selected bearing is a. Within.5 SD s of its mean value? (Φ(.5 Φ( b. Farther than 2.5 SD d from its mean value? ( [Φ(2.5 Φ( 2.5] c. Between and 2 SD s from its mean value? ([Φ(2 Φ(] [Φ( Φ( 2] The inside diameter of a randomly selected piston ring is a r.v. with mean value 2 cm and standard deviation 0.04 cm. a. If X is the sample of n 6 rings, where is the sampling distribution of X centred, and what is the standard deviation of the X distribution? X is centred at 2, V [ X] σ 2 /n ( /6 or, s.d. is 0.04/ b. wer the questions posed in part (a for a sample of size n 64 rings. X is centred at 2, V [ X] σ 2 /n ( /64 or, s.d. is 0.04/ c. For which of the two random samples, the one of part (a, or the one of part (b, is X more likely to be within 0.0 cm of 2 cm? Eplain your reasoning. part (b, because, (bhas smaller variance/s.d. 8

9 3. The lifetime of a certain type of battery is normally distributed with mean value 8 hours and standard deviation hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package eceeds that value for only 5% of all packages? Individual lifetimes be denoted by X i N(µ, σ 2. Then the total lifetime of 4 batteries is given by Y X X 2 X 3 X 4. Therefore, Y N(4µ, 4σ 2 or Y N(4 8, 4 2 or Y N(32, 4. Therefore, [Y 32]/2 N(0,. Let z 0 be the required critical point, then z0 32 Φ We know that Φ( Therefore, z z Each of 50 newly manufactured items is eamined and the number of scratches per item is recorded (the items are supposed to be free of scratches, yielding the following data: Scratches/item Observed freq Let X the number of scratches on a randomly selected item and assume that X has a Poisson distribution with parameter λ (P (X k ep( λλ k /k!. Find an unbiased estimator of λ and compute the estimate for the above data. [Hint: E(X λ If E[X] λ, implies E[ X E[ n X i ] n i E[X i ] i [nλ]/n λ Therefore X is an unbiased estimator of λ. The estimate is simply, the mean of the above data, ( (

10 5. A random sample of n bike helmets manufactured by a certain company is selected. Let X the number among the n that are flawed and let p P (flawed. Assume that only X is observed, rather than the sequence of Successes and Failures. Derive the Maimum Likelihood Estimator of p. If n 20 and 3, what is the estimate? Is this estimator unbiased? This is a Binomial model. Therefore, p.d.f. is n P [X ] p ( p n Because we have just a single observation, the jt. p.d.f itself is this, and probability in this case is not 0, we can take log, n ln P [X ] ln[ ] ln p (n ln( p Maimizing w.r.t. p, ln P [X ] p p n p ( 0 (n p ( p np ˆp n To show unbiasedness, use the fact that X is binomial, E[X/n] (/ne[x] np/n p The following page contains a table of cumulative standard normal probabilities Φ( P [Z < ] e 2 2 d 2π For < 0, Φ( Φ(. 0

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