STAT 3502
Solution - Assignment # 2 Total mark=100
1. A large industrial firm purchases several new word processors at the end of each year, the exact number depending on the frequency of repairs in the previous year. Suppose that the number of word processors, X, purchased each year has the following proba-bility distribution
x 0 1 2 3
probability 1/10 3/10 2/5 1/5
a. If the cost of desires model is $1200 per unit and at the end of the year a refund of 50X2 dollars will be issued, how much can this firm expect to spend a new word processors during this year?
b. Find expected value and variance for 2X2 − 4.
Sol:
a) [5] The expect money to spend=E(1200X) − E(50X2) = 1200E(X) − 50E(X2)
where E(X) = 0(1/10) + 1(3/10) + 2(2/5) + 3(1/5) = 17/10 and E(X2) = 0(1/10) +
1(3/10) + 4(2/5) + 9(1/5) = 37/10 so
E(1200X) − E(50X2) = 1200(17/10) − 50(37/10) = 1885
b) [4] E(2X2− 4) = 2E(X2) − 4 = 2(37/5) − 4 = 10.8,
V (2X2− 4) = 4V (X2) = 4(E(X4) − (E(X2))2), where E(X4) = 0(1/10) + 1(3/10) +
16(2/5) + 81(1/5) = 22.9, so V (X) = 4(22.9 − (3.7)2) = 36.84
2. The density function of the continuous random variable X, the total number of hours in units of 100 hours that a family runs a vacuum cleaner over a period of one year is
f (x) =      x 0 < x < 1 2 − x 1 ≤ x < 2 0 otherwise.
Find the average number of hours per year that families run their vacuum cleaner. Sol: [8]
The average number of hours per year=100(E(X)) = 100(R1
0 x
2dx +R2
1 x(2 − x)dx) =
100([x3/3]1
0+ [x2+ x3/3]21) = 100(1) = 100
3. Hospital administrators in large cities anguish about traffic in emergency rooms. At a particular hospital in a large city, the staff on hand cannot accommodate the patient traffic if there are more than 10 emergency cases in a given hour. It is assumed that patient arrival follows a Poisson process, and historical data suggest that, on the average, 5 emergencies arrive per hour.
a. What is the probability that in a given hour the staff cannot accommodate the patient traffic?
b. What is the probability that more than 20 emergencies arrive during a 3-hour shift? Sol:
a) [3] λ = 5 per hour, so P (X > 10) = 1 − P (x ≤ 10) = 1 − 0.9863 = 0.0137 b) [3] λ = 15 per 3 hour, so P (X > 20) = 1 − P (x ≤ 20) = 1 − 0.917 = 0.083
4. A roofing contractor offers an inexpensive method of fixing leaky roofs. However, the method is not foolproof. The contractor estimates that after the job is done, 10 percent of the roofs will still leak.
a. The resident manager of an apartment complex hire this contractor to fix the roof a six building. What is the probability that after the work has been done, at least two of these roofs will still leak?
b. The contractor offers a guarantee that pays $100 if a roof still leaks after he has worked on it. If he works on 81 roofs, find the mean and standard deviation of the amount of money that he will have to pay as a results of this guarantee.:
Sol:
a) [4] X follows Binomial distribution with n = 6, p = 0.1, so P (X ≥ 2) = 1 − [P (X = 0) + P (X = 1)] = 1 − (C6 0(0.1)0(0.9)6+ C16(0.1)1(0.9)5) == 1 − 0.8857 = 0.1143 b) [4] n = 81, p = 0.1, E(100X) = 100E(X) = 100(81 ∗ 0.1) = 810, σ100X = 100 √ 81 ∗ 0.1 ∗ 0.9 = 270
5. A lawyer commutes daily from his suburban home to his midtown office the average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of the trip time to be normally distributed.
a. What is the probability that a trip will take at least 1/2 hour?
b. If the office opens at 9:00am and the lawyer leaves his house at 8:45am daily, what percentage of the time is he late for work?
c. If he leaves the house at 8:35am and coffee is served at the office from 8:50am until 9:00am. What is the probability the has misses coffee?
d. Find the length of time above which find the slowest 15% of trips.
e. Find the probability that 2 of the next 3 trips will take at least 1/2 hour? Sol: a) [3] X ∼ N (24, (3.8)2), so P (X > 30) = 1 − P (Z < 30−243.8 ) = 1 − P (Z < 1.58) = 0.0571 b) [3] P (X > 15) = 1 − P (Z < 15−243.8 ) = 1 − P (Z < −2.37) = 0.9911, he will be late in 99.11% of the time. c) [4] P (X > 25) = 1 − P (Z < 25−24 3.8 ) = 1 − P (Z < 0.26) = 0.3974
d) [4] P (X < x) = 0.15, or P (Z < z) = 0.15 → z w 1.04, then x = 3.8(1.04) + 24 = 27.952
e) [4] Y : number of trips which take at least 1/2 hour, then Y ∼ Bin(3, 0.057), and P (Y = 2) = C3
2(0.057)2(1 − 0.057) = 0.0092
6. In an assembly-line production of industrial robots, gearbox assembles can be installed in 1 minute each if holes have been properly drilled in the boxes and in 10 minutes if the holes must be redrilled. Twenty gearboxes are in stock, 2 with improperly drilled holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots.
a. Find the probability that all five gearboxes will fit properly.
b. Find the mean, variance and standard deviation of the time it takes to install these five gearboxes.
Sol:
a) [4] Denote by X the number of improperly drilled gearboxes in the sample of 5, then X follows a hypergeometric distribution with n = 5, N = 20, and M = 2
P (X = 0) = ( 10 5)( 2 0) (20 5) = 0.553
b) [5] The total time, T , that it takes to install the boxes is T = 10X + (5 − X) = 9X + 5 First,E(X) = nMN = 5(2/20) = 0.5, V ar(X) = nMN(1 − MN)(N −nN −1) = 5(0.1)(1 − 0.1)(15/19) = 0.322 so E(T ) = 9E(X) + 5 = 9.5, V ar(T ) = 92V ar(X) = 81(0.355) = 28.755, σT = √ 28.755 = 5.362
7. The percentage of impurities per batch in a chemical product is a random variable X with pdf
f (x) = (
12x2(1 − x) 0 ≤ x ≤ 1
0 otherwise.
A batch with more than 40% impurities cannot be sold.
a. Find the probability that a randomly selected batch cannot be sold because of excessive impurities. b. Find the cdf, F (x). c. Use F (x) to find P (0.5 ≤ x < 0.8). Sol: a) [4] P (X > 0.4) =R0.41 (12x2− 12x3)dx = (4x3− 3x4)|1 0.4 = 4 − 3 − (0.256 − 0.0768) =
0.8208 b) [5] By definition of cd F (x) F (x) = Z x −∞ f (t)dt =      0 x ≤ 0 Rx 0(12y 2− 12y3)dy 0 < x < 1 1 x ≥ 1 =      0 x ≤ 0 4x3− 3x4 0 < x < 1 1 x ≥ 1 c) [3] P (0.5 ≤ x ≤ 0.8) = F (0.8) − F (0.5) = 4(0.8)3− 3(0.8)4 − 4(0.5)3+ 3(0.5)4 = 0.5067
8. The number of flaws on a VHS magnetic tape produced continuously at a factory fol-lows a Poisson distribution with an average of 0.01 flaws per meter. A standard VHS cassette tape contains 250 meters of magnetic tape.
a. What is the probability that there are at least two flaws in a single VHS cassette tape?
b. What is the probability that there are no flaws in a single VHS cassette tape; that is, a tape is flawless?
c. In a random sample of 20 cassettes, what is the probability that at least three of them are flawless?
Sol:
Let X represent the number of flaws in single VHS cassette tape, then X ∼ P (λ = 0.01 × 250 = 2.5)
a) [3] P (X > 2) = 1 − [P (X = 0) + P (X = 1)] = 1 −he−2.50!2.50 +e−2.51!2.51i= 0.7127 b) [4] P (X = 0) = e−2.50!2.50 = 0.0821
c) [4] Let Y represent the number of flawless VHS cassette tapes out of the 20 tapes. Then Y ∼ B(20, 0.0821). P (Y > 3) = 1 − [P (Y = 0) + P (Y = 1) + P (Y = 2)] 1 − 200(0.0821)00.917920+ 20 1(0.0821) 10.917919+ 20 2(0.0821) 20.917918 = 0.2233
9. For each of the following determine the value of c for which p(x) is a pmf: a. p(x) = c(1/6)x, x = 1, 2, 3, · · ·
b. p(x) = cx, x = 1, 2, 3, · · · , n
Sol:
The value of constant c is found from the condition P
all xp(x) = 1
a) [4] Using the geometric formula
∞
X
k=0
qk = 1
we get 1 = c ∞ X x=1 (1/6)x = c ∞ X x=0 (1/6)x− 1 ! = c 1 1 − 1/6 − 1 = c 5 hence c = 5. b) [5] Recall that n X k=1 k = n(n + 1) 2 therefore c n X x=1 n(n + 1) 2 = 1 and hence c = n(n+1)2 .
10. Twenty percent of all cars manufactured by a certain company have a defective trans-mission system. If a dealer has sold 200 of these cars, what is the probability that it will need to service at most 50 of them?
Sol: [5]
p = 0.2, n = 200, so np = 40 > 10, npq = 32 > 10 P (X 6 50) = P (Z 6 50−40+0.5√
32 ) = P (Z < 1.86) = 0.9686
11. A machine to fill cereal boxes produces fills (in ounces) that are normally distributed. Suppose that its standard deviation σ can be carefully adjusted. What is the largest value of σ that will allow the actual value dispensed to be within one ounce of the mean with probability at least 0.95?
Sol:[5] X ∼ N (µ, σ2) P (µ − 1 6 X 6 µ + 1) > 0.95 → P (µ−1−µσ 6 Z 6 µ+1−µ σ ) = P ( −1 σ 6 Z 6 1 σ) > 0.95 → Φ(1 σ) − Φ( −1 σ ) = 2Φ( 1 σ) > 0.95 → Φ( 1 σ) > 0.975 → 1 σ > 1.96 → σ 6 0.51
