Statistics and Probability_Chapter 2_FTU

CHAPTER 2. DISCRETE RANDOM VARIABLES
#1/5. For each of the following, indicate if a discrete or a continuous random variable
provides the best definition:
a. The time to write a statistics exam in a university where the time limit is 3 hours
and students cannot leave before 30 minutes. → A continuous random variable
b. The number of cars that arrive each day for repair in a two-person repair shop. →
A discrete random variable
c. Total daily e-commerce sales in dollars. → A continuous random variable
d. The gender of a person that we observe at random in a street. → IS NOT random
variable
e. The number of defective items in a sample of 20 items from a large shipment. →
A discrete random variable
#2/9. Show the probability distribution function (distribution table) of the number of
heads when two fair coins are tossed independently.
 X = number of heads when tossing two coins
 Possible values: HH, HT, TH, TT
X 0 1 2
P 1/4 1/2 1/4
#3/9. A box has 10 products, including 4 defective products. Pick 2 products at
random. Make a probability distribution table of the number of defective
products taken out?

The total number of ways to pick 2 products out of 10:
Total ways = 𝐶2
10
=
10!
2!(10−2)!
= 45 = 𝑁
• Case 1: X = 0 (No defective products)
o Total outcomes = 𝐶0
4
× 𝐶2
6
= 15
o 𝑃(𝑐𝑎𝑠𝑒 1) =
15
45
=
1
3
• Case 2: X = 1 (1 defective product)
4
× 𝐶1
6
= 24
24
45
=
8
15
• Case 3: X = 2 (2 defective products)
4
× 𝐶0
6
= 6
6
45
=
2
15
The probability distribution table
Number of Defective Products (X) Probability P(X)
0 1/3
1 8/15
2 2/15
#4/9. There are 2 packages. Package 1 has 3 good products and 2 bad products.
Package 2 has 2 good products and 3 bad products. Randomly take 2 products from
package 1 and 1 product from package 2. Establish a distribution table of the
number of good products among the 3 selected products?
Package 1
• Good products: 3
• Bad products: 2

• Total products: 5
 Selecting 2 products
Total ways to pick 2 products from package 1 : 𝑁1 = 𝐶2
5
= 10
*Case 1 (X = 0): 𝑃 =
𝐶0
3×𝐶2
2
10
=
1
10
*Case 2 (X = 1): 𝑃 =
𝐶1
3×𝐶1
2
10
=
3
5
*Case 3 (X = 2): 𝑃 =
𝐶2
3×𝐶0
2
10
=
3
10
Package 2
• Good products: 2
• Bad products: 3
• Total products: 5
 Slecting 1 product
Total ways to pick 2 products from package 1 : 𝑁1 = 𝐶1
5
= 5
*Case 1 (X = 0): 𝑃 =
𝐶0
2×𝐶1
3
10
=
3
5
*Case 2 (X = 1): 𝑃 =
𝐶1
2×𝐶0
3
10
=
2
5
Total selected products: 3
Possible number of good products (X): 0, 1, 2, 3
The probability distribution table
Number of Good Selected Products (X) Probability P(X)
0 1
10
×
3
5
=
3
50

1
(
1
10
×
2
5
) + (
3
5
×
3
5
) =
2
50
+
18
50
=
2
5
2
(
3
5
×
2
5
) + (
3
10
×
3
5
) =
12
50
+
9
50
=
21
50
3 3
10
×
2
5
=
3
25
#7/18. A contractor is interested in the total cost of a project on which she intends to
bid. She estimates that materials will cost $ 25,000 and that her labor will be $900
per day. The project will takes X days to complete. Given the probability distribution
for completion times.
X 10 11 12 13 14
P 0.1 0.3 0.3 0.2 0.1
a. Find the mean and variance for completion time X
The mean: 𝜇 = ∑ 𝑋 × 𝑃(𝑋)
= 10 × 0.1 + 11 × 0.3 + 12 × 0.3 + 13 × 0.2 + 14 × 0.1 = 11.9
The variance: 𝜎2
= ∑(𝑋 − 𝜇)2
× 𝑃(𝑋)
= (10 − 11.9)2
× 0.1 + (11 − 11.9)2
× 0.3 + (12 − 11.9)2
× 0.3
+ (13 − 11.9)2
× 0.2 + (14 − 11.9)2
× 0.1 = 1.29
b. Find the mean, variance and standard deviation for total cost C
*Method 1: The probability distribution of total cost C: 𝑌 = 25,000 + 900𝑋
Y 34,000 34,900 35,800 36,700 37,600
P(Y) 0.1 0.3 0.3 0.2 0.1
(𝑌 − 𝜇)2 2,924,100 656,100 8,100 980,100 3,572,100

The mean: 𝜇 = 34,000 × 0.1 + 34,900 × 0.3 + 35,800 × 0.3 + 36,700 × 0.2 +
37,600 × 0.1 = 35,710
The variance: 𝜎2
= 2,924,100 × 0.1 + 656,100 × 0.3 + 8,100 × 0.3 +
980,100 × 0.2 + 3,572,100 × 0.1 = 1,044,900
The standard deviation: 𝜎 = √1,044,900 = 90√129 = 1022.203502
*Method 2:
𝐸(𝑌) = 𝜇𝑌 = 900𝐸(𝑋) + 25,000 = 900 × 11.9 + 25,000 = 35,710
𝜎𝑌
2
= 9002
× 𝜎𝑋
2
= 9002
× 1.29 = 1,044,900
#8/20. Given the probability of distribution function:
a. Calculte the cumulative probability distribution
X 0 1 2
P(X) 0.25 0.5 0.25
(𝑋 − 𝜇)2 1 0 1
b. Find the mean, variance and standard deviation
𝐸(𝑋) = 𝜇 = 0 × 0.25 + 1 × 0.5 + 2 × 0.25 = 1
𝜎2
= 𝑉(𝑋) = 1 × 0.25 + 0 × 0.5 + 1 × 0.25 = 0.5
𝜎 = √0.5 =
√2
2
#9/20. A shipment of 20 parts contains 2 defectives. Two parts are chosen at random
from the shipment and checked. Let the random variable Y denote the number of

defectives found. Find the probability distribution of this random variable. Find the
mean and variance of the random variable Y.
Probability distribution table
Y 0 1 2
P(Y) 𝐶2
18
𝐶2
20 = 0.805
𝐶1
18
× 𝐶1
2
𝐶2
20 = 0.189
𝐶2
2
𝐶2
20 = 0.005
𝐸(𝑌) = 𝜇 = 0 × 0.805 + 1 × 0.189 + 2 × 0.005 = 0.199
𝑉(𝑌) = 𝜎2
= (0 − 0.199)2
× 0 + (1 − 0.199)2
× 0.189 + (2 − 0.199)2
× 0.005
= 0.169
#10/20. Avery large shipment of parts contains 10% defectives. Two parts are chosen
at random from the shipment and checked. Let the random variable X denote the
number of defectives found. Find the probability distribution of this random variable.
Find the mean and variance of the random variable X.
Probability distribution table
X 0 1 2
P(X) 0.9 × 0.9 = 0.81 0.1 × 0.9
+ 0.1 × 0.9
= 0.18
0.1 × 0.1 = 0.01
𝐸(𝑋) = 𝜇 = 0 × 0.81 + 1 × 0.18 + 2 × 0.001 = 0.2
𝑉(𝑋) = 𝜎2
= (0 − 0.2)2
× 0.81 + (1 − 0.2)2
× 0.18 + (2 − 0.2)2
× 0.01 = 0.18
#11/21

Statistics and Probability_Chapter 2_FTU

Binomal distribution_#13/25. Suppose that you are in charge of marketing airline
seats for a major carrier. Four days before the flight date you have 16 seats remaining
on the plane. You know from past experience data that 80% of the people that
purchase tickets in this time period will actually show up for the flight.
a. If you sell 20 extra tickets, what is the probability that you will overbook the
flight?
b. If you sell 18 extra tickets, what is the probability that you have at least 1
empty seat?
Binomal distribution_#14/26. Aproduction manager knows that 5% of components
produced by a particular manufacturing process have some defect. Six of these
components, whose characteristics can be assumed to be independent of each other,
are examined.
a. What is the probability that none of these components has a defect?
n = 6; p = 0.05; x = 0

The probability that no element has a defect
𝑃(𝑥) = 𝐶𝑥
𝑛
× 𝑝𝑥(1 − 𝑝)𝑛−𝑥
⇔ 𝑃(0) = 𝐶0
6
× 𝑝0(1 − 0.05)6−0
= 0.7531
b. What is the probability that one of these components has a defect?
The probability if one component is defective
𝑃(𝑥) = 𝐶𝑥
𝑛
× 𝑝𝑥(1 − 𝑝)𝑛−𝑥
⇔ 𝑃(1) = 𝐶1
6
× 𝑝1(1 − 0.05)6−1
= 0.2321
c. What is the probability that at least two of these components have a defect?
The probability that there are at least 2 components with a defect.
𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 < 2) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)]
= 1 − 0.7531 − 0.2321 = 0.0148
Binomal distribution_#15/27. A notebook computer dealer mounts a new
promotional campaign. Purchasers of new computers may, if dissatisfied for any
reason, return them within 2 days of purchase and receive a full refund. The cost to
the dealer of such a refund is $100. The dealer estimates that 15% of all purchasers
will, indeed, return computers and obtain refunds. Suppose that 50 computers are
purchased during the campaign period.

a. Find the mean and standard deviation of the number of these computers that
will be returned for refunds.
𝑛 = 50; 𝑝 = 0.15; 𝐸(𝑥) = 𝑛 × 𝑝 = 50 × 0.15 = 7.5 𝜇 = 𝑛 × 𝑝(1 − 𝑝)
𝜎 = √𝑝(1 − 𝑝) × 𝑛 = √0.15(1 − 0.15) × 50 = √6.375
b. Find the mean and standard deviation of the total refund costs that will
accrue as a result of these 50 purchases.
𝑛 = 100; 𝑝 = 7.5; 𝐸(𝑥) = 𝑛 × 𝑝 = 100 × 7.5 = 75
𝜎𝑡𝑜𝑡𝑎𝑙 = 𝜎 × 𝑛 = √6.375 × 100 = 25√102
Poisson distribution_#16/30. Andrew Whittaker, computer center manager, reports
that his computer system experienced three component failures during the past 100
days.
a. What is the probability of no failures in a given day?
The expected number of failures per day is 𝜆 =
3
100
= 0.03
𝑃(𝑛𝑜 𝑓𝑎𝑖𝑙𝑢𝑟𝑒𝑠) = 𝑃(𝑋 = 0) =
𝑒−0.03
× 0.030
0!
= 0.97
b. What is the probability of one or more component failures in a given day?
𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑓𝑎𝑖𝑙𝑢𝑟𝑒) = 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1)
= 1 − 𝑃(𝑋 = 0) = 1 − 0.97 = 0.029
c. What is the probability of at least two failures in a 3-day period?

Poisson distribution_#17/31. Customers arrive at a busy checkout counter at an
average rate of 3 per minute. If the distribution of arrivals is Poisson, find the
probability that in any given minute there will be 2 or fewer arrivals.
- The average arrival rate is 𝜆 = 3 𝑝𝑒𝑟 𝑚𝑖𝑛𝑢𝑡𝑒
- The distribution of arrivals is Poisson
𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)
=
𝑒−3
× 30
0!
+
𝑒−3
× 31
1!
+
𝑒−3
× 32
2!
= 𝑒−3
+ 3𝑒−3
+
9
2
𝑒−3
=
17
2
𝑒−3
The probability that there will be 2 or fewer arrivals in any given minute is
approximately (0.4232).
Poisson distribtuion #18/31. The number of accidents in a production facility has a
Poisson distribution with a mean of 31.2 per month.
a. For a given month what is the probability there will be fewer than 2 accidents?
𝜆 = 31.2 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ
𝑃(𝑋 < 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) =
𝑒−31.2
× 31.20
0!
+
𝑒−31.2
× 31.21
1!
= 𝑒−31.2
+ 31.2𝑒−31.2
= 32.2𝑒−31.2
b. For a given month what is the probability there will be more than 3 accidents?

(𝑋 > 3) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)]
=
𝑒−31.2
× 31.20
0!
+
𝑒−31.2
× 31.21
1!
+
𝑒−31.2
× 31.22
2!
+
𝑒−31.2
× 31.23
3!
= 𝑒−31.2
+ 31.2𝑒−31.2
+ 486.72𝑒−31.2
+ 5061.888𝑒−31.2
= 5580.808𝑒−31.2

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