Review

1
Review 1

• Chapter 2
• Chapter 3
• Chapter 4

2
Chapter 2 Descriptive Statistics Numerical
Methods

• Problem 1 (Calculations)
• The ages of employees of a fast-food outlet are
  as follows 19, 19, 65, 20, 21, 18, 20.
• a) Compute the mean, the median, and the mode of
  the ages
• Mean (191920)/7 26
• Median the center of the sorted series 20
  18, 19, 19, 20, 20, 21, 65
• Mode the number with the largest frequency
  19, 20

Mean 19.5
Median 19.5 (1920)/2
Mode 19, 20 no change.
b) Assume the oldest employee retires
The mean is sensitive to extreme value the
median and the mode are less sensitive.
3
Measures of Central Tendency (Excel,
interpretation)

• Problem 2 (Excel, interpretation)
• The summer income of a sample of 125 second-year
  business students are stored in Prob 2
• Calculate the mean and the median
• What do the two measures of central location tell
  you about the income?
• Which measure should be used to summarize the
  data?

4
Measures of Central Tendency

• Problem 2 - solution
• The distribution is reasonably symmetrical, but a
  few low incomes have pulled the mean below the
  median, resulting in a distribution slightly
  skewed to the left.
• Either measure could be used, but the median is
  better because it is not affected by a few low
  incomes.

5
Measures of Central Tendency

• Problem 2 - solution
• The distribution is reasonably symmetrical, but a
  few low incomes have pulled the mean below the
  median, resulting in a distribution slightly
  skewed to the left.
• Either measure could be used, but the median is
  better because it is not affected by a few low
  incomes.

6
Measures of Central Tendency

• Problem 3 (Excel, interpretation)
• The owner of a hardware store that sells
  electrical wires by the meter is considering
  selling the wire in pre-cut lengths to save on
  labor cost. A sample of wire sold over the course
  of 1 week was recorded (Prob3.xls).
• A) Compute the mean, median and mode
• B) What is the weakness of each measure in
  providing useful info?
• C) How might the owner decide on the lengths to
  pre-cut?

7
Measures of Central Tendency

• A) The mean resides to the right of the median.
  The distribution of lengths is somewhat
  asymmetrical, skewed to the right (there must be
  some long wires sold that affect the mean value).

8
Measures of Central Tendency

• B) The mean is unduly influenced by extreme
  observations.
• The median doesnt indicate what lengths are
  most preferred.
• The mode doesnt consider any desired lengths
  other than the one most frequently purchased.

9
Measures of Central Tendency

• C) You may draw the cumulative distribution and
  trim the tails

3
10
4
11
10
Measures of Variability

• Problem 4Calculate the mean, variance and the
  standard deviation of the following set of
  numbers, treating them as
• Sample
• Population
• The set is 14, 7, 8, 11, 5

For the standard deviation take the square root
of the variance
11
The variance

• Problem 5 (The variance, Excel)
• The number of customers entering a bank each hour
  for the last 100 days was recorded (Problem 5).
• For each hour determine the mean and standard
  deviation.
• What do these statistics tell you?

12
The Variance

• Problem 5 solution
• The noon hour (121) is the busiest, followed by
  the (23 P.M.) and (1011 A.M.) periods.
• The variances during the noon hour and between
  10-11 AM are the largest, which makes it
  difficult to predict the number of customer
  entering the bank.
• Staff lunch breaks and coffee breaks should be
  scheduled with this in mind.

Comment All the samples can be analyzed in a
single run by Excel gt Descriptive statistics.
13
The Variance
14
The Variance
69 7 62 69 7 76
Problem 5 - solution
a)

• b) For a mound shaped distribution the Empirical
  Rule applies. Thus,
• Approximately (.68)(500) 340 grades are in
  (62, 76)
• Approximately (.95)(500) 475 grades are in
  (55, 83)
• Virtually all of the grades are in (48, 90)

15
Chebychev Theorem

• If the distribution is not mound shaped we need
  to use Chebychev Theorem

c) If the distribution is not mound shaped, at
least (8/9)(500) 444.4 (or 445) grades are
within (48, 90)
16
Chapter 3 (Probability)

• Probability is a numerical measure that
  represents the likelihood of occurrence of a
  random event.
• 0P(A) 1
• Relationships between events
• Union Event A or event B have occurred (at least
  one of them took place).
• Intersection Event A and event B have occurred
  (both event took place simultaneously).
• Complement event If event A did not occur, then
  event called Not A (A) occurred.

17
Chapter 3

• Problem 6
• A firm classifies its customers accounts in two
  ways By balance and whether it is overdue.
• Account balance Overdue Not OverdueUnder
  100 .08 .42100 - 500 .08 .22Over
  500 .04 .16
• Define the following events
• A An account is under 100
• B An account is overdue
• An account is selected at random

Find the following probabilities
P(under 100)P(A) .08.42.50
P(under 100 and overdue) P(A and B).08.
18
Chapter 3

• Problem 6
• A firm classifies its customers accounts in two
  ways By balance and whether it is overdue
• Account balance Overdue Not OverdueUnder
  100 .08 .42100 - 500 .08 .22Over
  500 .04 .16
• Define the following events
• A An account is under 100
• B An account is overdue

Find the following probabilities
P(under 100 or overdue)
P(A or B) .08.42 .08.08.04.70
19
Chapter 3

• Problem 6.11 (Relationshipsand, or, conditional)
• A firm classifies its customers accounts in two
  ways By balance and whether it is overdue
• Account balance Overdue Not OverdueUnder
  100 .08 .42100 - 500 .08 .22Over
  500 .04 .16
• Define the following events
• A An account is under 100
• B An account is overdue

Find the following probabilities
P(under 100 or overdue) P(A or B) .08.42
.08.08.04.62
P(not overdue)P(not B) .42.22.16.80.
Or, P(not B) 1 P(B) 1 (.08.08.04) 1
- .20.
More events and their probabilities
20
Chapter 3
Note P(A).50 but P(AB).40

• Problem 7 (conditional probability)
• A firm classifies its customers accounts in two
  ways By balance and whether it is overdue
• Account balance Overdue Not OverdueUnder
  100 .08 .42100 - 500 .08 .22Over
  500 .04 .16
• Define the following events
• A An account is under 100
• B An account is overdue

Find the following probabilities

• If the account selected is overdue, what is
  the probability that its balance is under
  100? That is
• P(AB)?

P(AB).08/(.08.08.04)
.08/(.20).40
P(AB)P(A and B)/P(B)
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Chapter 3

• Problem 7
• A firm classifies its customers accounts in two
  ways By balance and whether it is overdue.
• Account balance Overdue Not OverdueUnder
  100 .08 .42100 - 500 .08 .22Over
  500 .04 .16
• Define the following events
• A An account is under 100
• B An account is overdue

Find the following probabilities
If the account selected is overdue what is the
probability that its balance is 500 or less?
P(CD)P(C and D)/P(D)
P(500 or lessOverdue) P(500 or less, and
Overdue)/P(Overdue) (.08.08)/(.08.08.04).80
22
Chapter 3

• Problem 8 (Multiplication rule)
• Sporting goods store estimates that 20 of the
  students at a nearby university ski downhill, and
  15 ski cross-country. Of those who ski downhill,
  40 also ski cross-country.
• What percentage of the students ski both
  downhill and cross-country?

Define events A a student ski downhill B a
student ski cross-country
Given probabilitiesP(A) .2 P(B) .15
P(BA) .4
Calculate P(A and B) P(BA)P(A) (.4)(.2)
.08
23
Chapter 3

• Problem 9 (Addition rule)
• Sporting goods store estimates that 20 of the
  students at a nearby university ski downhill, and
  15 ski cross-country. Of those who ski downhill,
  40 also ski cross-country.
• What percentage of the students do not ski at
  all?

Calculate P(not A and not B) 1 P(A or B) P(A
or B) P(A) P(B) P(A and B) (.2) (.15)
(.08) .27 Therefore P(not A and not B) 1
P(A or B) 1-.27 .73
24
Chapter 3

• Problem 10 (Independent events, multiplication
  rule)
• Approx. 3 out of every 4 Americans received a
  refund from the IRS in 1995. If 3 individuals
  are selected at random find the probabilities of
  the following events
• All three received a refund
• None received a refund
• At least one received a refund
• Exactly one received a refund

25
Chapter 3

• Problem 10 solution
• Let A be the event Individual 1 received a
  refund. Define B and C similarly for individual 2
  and 3. Then P(A)P(B)P(C)3/4.
• P(All the three received a refund)P(A and B and
  C)P(A)P(B)P(C)(3/4)3
• P(None received a refund)P(Not A, and not B, and
  not C) P(not A)P(not B)P(not C)(1/4)3
• P(At least one received)1-P(none
  received)1-(1/4)3.
• P(Exactly one received a refund)
• P(A and not B and not C)
• P(not A and B and not C)
• P(not A and not B and C)

(3/4)(1/4)(1/4)
(1/4)(3/4)(1/4)
(1/4)(1/4)(3/4)
3(3/4)(1/4)2.
26
Chapter 4Random Variables

• Problem 11 (discrete random variable, expected
  value, variance)
• You and a friend have contributed equally to a
  portfolio of 500. The annual income (X) has the
  following distribution x 500 1,000 2,000
• P(x) .5 .3 .2
• Determine the annual expected value and variance
  of the income earned on this portfolio.
• Determine the net annual profit and variance to
  you.
• What is the expected profit and variance to you
  for the next two years?

27
Chapter 4Random Variables

• Problem 11 solution
• E(X)(500)(.5)(1000)(.3)(2000)(.2)
  950V(X)(500-950)2(.5)(1000-950)2(.3)(2000-
  950)2(.2)2 322500
• E(Ann. profit)E(X/2 250)E(X/2)-E(250)
  1/2E(X) 250 (½)950 250 225V(Ann.
  profit)V(X/2 250) V(X/2)V(250)
  V(X/2)0(1/2)2V(X)1/4(322500).

28
Chapter 4Random Variables

• Problem 11 solution continued
• If Xi is the income for year i, the income for
  the next two years is X1 X2. Your profit is
  therefore, (½)(X1 X2) 250.
• E(2 years Profit) E(½X1½X2 250) ½ E(X1)
  ½ E(X2) 250 assuming the income distribution
  does not change between the two years (½) 950
  (½)950 250 700.
• V(2 years Profit) V(½X1½X2 250) assuming
  the income distribution does not change between
  the two years, and the incomes in the two years
  are independent random variables
  (½)2V(X1)(½)2V(X2)

29
Chapter 4The Binomial Distribution

• Example 12
• A survey reported that 20 of elementary school
  teachers use the Web. Fifteen teachers are
  selected at random. Answer the following
  questions.

30
Chapter 4The Binomial Distribution

• Solution Let us analyze this experiment first.
• There are n15 independent experiments.
• Each experiment has two possible outcomes.
• The probability of success in each experiment is
  p.20. which does not change from experiment to
  experiment.
• Therefore this is a binomial experiment.
• Define X the number of teachers that use the
  Web. X is binomial with parameters n15, and
  p.2.

31
Chapter 4The Binomial Distribution

• P(No teacher uses the Web)
• P(One teacher uses the Web)
• P( of Web users does not exceed 8)
• P(More than 2 Web users)P(X³3)P(X3)P(X4)P(
  X15)ltLet us use the binomial tablegt 1-
  P(X2)1-.398.602

Binomial table
32
Chapter 4The Binomial Distribution

• The expected number of teachers using the
  internet E(X)np15(.2)3 userThe variance
  of the number of Web users.V(X)np(1-p)15(.2)(.
  8)2.4 users2. Standard deviationV(X)1/2.
• P(Less than 8 are Web users, given that more than
  2 are users) P(X7X³3 P(X7 and X³3)/P(X³3)
  P(X3)P(X7)/P(X³3)ltLet us use the tablegt
• P(X3)P(X7) P(X7)-P(X2) .996 - .398
  .598P(X³3)1-P(X2)1-.398.602
• P(X7X³3) .598/.602 .993

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Chapter 4The Binomial Distribution

• Solution continued
• Repeat this problem assuming 50 teachers were
  sampled. Since there are no tables available for
  this number of repeated trials (n), well use
  Excel.
• P(X0).850
• P(X1)50(.2)(.8)49
• P(X4)0.018496 ltGo to Excel gt Type
  BINOMDIST(4,50,.2, True)

34
Chapter 4 The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table.P(Zgt1.7)

Normal Table
?
Z
0
1.7
35
Chapter 4 The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table.P(Zgt1.7)?

.5-.4554.0446
From the Normal Table
Z
0
1.7
36
Chapter 4 The normal distribution

• Problem 1
• Find the following probabilities using the normal
  table.P(Zgt .95)?

.5.32890.8289
Normal Table
0
-.95
37
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table. P(-1.14Z1.55)?

P (-1.14(ltZlt0)P(0ltZlt1.55)
-1.14 1.55
Normal Table
0
38
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table. P(-1.14Z1.55)?

P (-1.14ltZlt0)P(0ltZlt1.55)
.3729
-1.14
1.14
Normal Table
0
39
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table.P(-1.14 Z 1.55)

.9394
P (-1.14(ltZlt0)P(0ltZlt1.55)
.3729.4394 .8123
.8123
Normal Table
-1.14 1.55
0
40
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table. P(-2.97 Z -1.38)?

Normal Table
-2.97 - 1.38
0
41
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table. P(-2.97 Z -1.38)?

P(0ltZlt 2.97)-P(0ltZlt 1.38)
Normal Table
-2.97 - 1.38
2.97
0
42
Chapter 4The normal distribution

• Problem 1 (calculating normal probabilities)
• Find the following probabilities using the normal
  table. P(-2.97 Z -1.38)?

P(0ltZlt 2.97)-P(0ltZlt 1.38)
P(0ltZlt 2.97)-P(0ltZlt 1.38)
With Excel typenormsdist(-1.38)-normsdist(-2.97
)
Normal Table
-2.97 - 1.38
1.38
0
43
Chapter 4The normal distribution

• Problem 2 (application)
• Mensa is an organization whose member posses IQs
  in the top 2 of the population. IQ is normally
  distributed with a mean of 100 and standard
  deviation of 16.
• Questions
• What is the probability that a randomly selected
  person have an IQ of 140 or more?
• What minimum IQ qualifies a person to be admitted
  to Mensa?
• What is the probability that a randomly selected
  person from among Mensas members have an IQ of
  more than 140?

44
Chapter 4The normal distribution

• Question 1 What is the probability that a
  randomly selected person have an IQ of 140 or
  more?
• Answer Define X as the IQ level of a
  person.P(Xgt140)P(Zgt(140 100)/16)P(Zgt2.5)
  .5-.4938.0062. With Excel type
  
  1-normdist(140,100,16,True)

Normal Table
0
45
Chapter 4The normal distribution

• Question 2 What minimum IQ qualifies a person
  to be admitted to Mensa?
• Answer Define X as the IQ level of a person.
• For a Mensa member P(XgtX0).02
  P(Zgt(X0-100)/16).02If we define Z0(X0-100)/16
  then P(ZgtZ0).02Let us first find Z0 and then X0.

Finally, determine X0 by2.055(X0 100)/16
X0 1002.055(16) 132.88
Normal Table
2.055
0
46
Chapter 4The normal distribution

• Question 3 What is the probability that a
  randomly selected person from among members of
  Mensa have an IQ of more than 140?
• Answer Define X as the IQ level of a person.
• P(Xgt140Xgt132.88) P(Xgt140 and
  Xgt132.88)/P(Xgt132.88)
• For comparison we have seen that
  P(Xgt140)P(Zgt(140 100)/16)P(Zgt2.5).5- 4938
  .0062!!
• No surprise. Given that a person belongs to
  Mensathe probability his/her IQ gt 140 is much
  larger thanthis of a person from the general
  population.

P(Xgt140)/P(Xgt132.88)P(Zgt2.5)/.02.0062/.02.31
Normal Table