Chapter 03 - Random Variables - Statistics for Business | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

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Môn: Statistics for Business (BAO8OIU)

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Tác giả:

Profile Mai Nguyệt
International University
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Statistics for Business | Chapter 03: Random Variables
1
STATISTICS FOR BUSINESS [IUBA]
CHAPTER 03
RANDOM VARIABLES
PART 1
1. Random Variable X – Probability P(X) – Cumulative Distribution Function F(X)
Random Variable Probability Cumulative Distribution Function
X P(X) F(X)
0 0.1 0.1
1 0.2 0.3
2 0.3 0.6
3 0.2 0.8
4 0.1 0.9
5 0.1 1.0
1.0
P
(
x
)
≥
0
for
values
x
P
(
x
)
=
1

F
(
x
)
=
P
(
X
≤
x
)
=
P
(
x
)

2. Expected Value, Variance, and Standard Deviation of Random Variable X
Expected Value (Mean)
µ
=
E
(
X
)
=
∑
xP
(
x
)

Variance
σ
=
V
(
x
)
=
E
[
(
x
−
µ
)
]
=
∑
(
x
−
µ
)
P
(
x
)

Standard Deviation
σ
=
SD
(
X
)
=
V
(
X
)
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Statistics for Business | Chapter 03: Random Variables
2
CALCULATOR INSTRUCTIONS FOR STATISTICS
Note: This page is only relevant for CASIO scientific calculator FX-570ES
Computing Mean and Standard Deviation of Random Variable X.
(Chapter 03 | Random Variables)
Step 01: Press MODE 3: STAT 1: 1 - VAR
SHIFT + | [ ] MODE SET UP
REPLAY
4: STAT
1: ON (Frequency ?)
Step 02: Input the data into two columns
Step 03: Compute the expected value of the random variable X
No.01: Put the cursor in the and (n+1)th line FREQ column
No.02: Press + SHIFT 1 | [STAT]
5: VAR
2: 
[ = ]
Step 04: Compute the standard deviation of the random variable X
No.01: Return the table of data and put the cursor in the (n+1)th line
and . FREQ column
No.02: Press to delete the value of the expected value in step 03. [ DEL ]
(This step is very important for you to remember. If you do not delete
the expected value you had already computed, you will get the wrong
result in this computation)
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Statistics for Business | Chapter 03: Random Variables
3
No.03: Press SHIFT + | [ ] 1 STAT
5: VAR
3: 
[ = ]
Step 04: Compute the variance of the random variable X
We can get the value of the variance of the random variable X by using
the following formula:
(
)
=
[

( )]
Example | Part 1
PROBLEM:
The number of defects in a machine
-
made product is a random variable X with the
following probability distribution:
x
P(x)
0
0.1
1
0.2
2
0.3
3
0.3
4
0.1
a. Show that P(x) is a probability distribution.
b. Find the probability 
(
1 < ≤3
)
.
c. Find the probability
(
1 < ≤4
)
d. Find the cumulative distribution function F(x).
e. Find the expected value, variance, and standard deviation of the random
variable X.
SOLUTION:
a.
P
(
x
)

=
P
(
0
)
+
P
(
1
)
+
P
(
2
)
+
P
(
3
)
+
P
(
4
)
= 0.1 + 0.2 + 0.3 + 0.3 + 0.1 = 1.0
Hence, P(X) is a probability distribution
b.
P
(
1
<
≤
3
)
=
P
(
2
)
+
P
(
3
)
=
0
.
3
+
0
.
3
=
0
.
6
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Statistics for Business | Chapter 03: Random Variables
4
c.
P
(
1
<
≤
4
)
=
P
(
2
)
+
P
(
3
)
+
P
(
4
)
=
0
.
3
+
0
.
3
+
0
.
1
=
0
.
7
d.
x
P(x)
F(X
0
0.1
0.1
1
0.2
0.3
2
0.3
0.6
3
0.3
0.9
4
0.1
1.0
e.
General Method
(
)

(
)
(
−
)
(
)
0
0.1
0.0
0.441
1
0.2
0.2
0.242
2
0.3
0.6
0.003
3
0.3
0.9
0.243
4
0.1
0.4
0.361
2.1
1.29
Expected Value (Mean):
E
(
X
)
= xP
(
x
)
= 2.1

Variance:
V X
( )
=
(
x − µ
)
P
(
x
)
= 1.29

Standard Deviation:
SD X
( )
= V X
( )
= √1.29 ≈ 1.1358
Calculator Method
To compute the expected value, standard deviation, and variance by
using the calculator, we follow these steps:
Step 01: Press MODE 3: STAT 1: 1 - VAR
SHIFT + MODE | [SET UP]
REPLAY
4: STAT
1: ON (Frequency ?)
Step 02: Input the data into two columns
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Statistics for Business | Chapter 03: Random Variables
5
Step 03: Compute the
expected value
of the random variable X
No.01: Put the cursor in the and
6
th
line FREQ column
No.02: Press + SHIFT 1 | [ ] STAT
5: VAR
2: 
[ = ]
(The available result you can get is 2.1)
Step 04: Compute the standard deviation of the random variable X
No.01: Return the table of data and put the cursor
in the and .
6
th
line FREQ column
No.02: Press to delete the value of the [ DEL ]
expected value in step 03.
(This step is very important for you to
remember. If you do not delete the expected
value you had already computed, you will get
the wrong result in this computation)
No.03: Press + SHIFT 1 | [ ] STAT
5: VAR
3: 
[ = ]
(The available result you can get is
1.13578166
Step 04: Compute the variance of the random variable X
Since
 
( )
=
( )
or
(
)
=
[ ( )]
, we can get the
result as following
(

)
=
[ ( ) (]
= 1.135781669
)
= . 
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Statistics for Business | Chapter 03: Random Variables
6
PART 2
1. Binomial Distribution: ~ (,)
Let be the number of trials
 be the probability of success
Conditions for a Binominal Random Variable
The trials must be Bernoulli trials in that the outcomes can only be either success or failure.
The outcomes of the trials must be independent.
The probability of success in each trial .
(
=
)
=
(
−
)
(
)
=
,
,
…
,
(
)
=

(
)
=

(
−
)
Example | Part 2.1: (Case of Binomial Distribution)
PROBLEM:
A salesperson goes door-to-door in a residential area to demonstrate the use of a new
household appliance to potential customers. At the end of a demonstration, the probability that
the potential customer would place an order for the product is a constant 0.2107. To perform
satisfactorily on the job, the salesperson needs at least four orders. Assume that each
demonstration is a Bernoulli trial.
a. If the salesperson makes 15 demonstrations, what is the probability that there would be
exactly 4 orders?
SOLUTION
With  = 0.2107 and  = 15, the probability that there would be EXACTLY 4 orders:
(
= 4
)
=
15
4
×
(
0.2107
)
×
(
1 − 0.2107
)

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Statistics for Business | Chapter 03: Random Variables
7
With the calculator, you type as following
( ( )

)
× .
×
( )
− .

After pressing , we can get the result [ = ] .
b. If the salesperson makes 16 demonstrations, what is the probability that there would be at
most 4 orders?
SOLUTION
With  = 0.2107 and  = 16, the probability that there would be AT MOST 4 orders:
(
≤4
)
=
(
0
)
+
(
1
)
+
(
2
)
+
(
3
)
+
( )
4
=
16
×
(
0.2107
)
×
(
1 − 0.2107
)


With the calculator, you use the following method
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type

( )
×
(
.
)
×
(
− .
)


Note: Press to get sign X in the formula + ALPHA [ ) ] | [ ]X
Step 03: Press we can get the result , [ = ] .
c. If the salesperson makes 17 demonstrations, what is the probability that there would be at
least 4 orders?
SOLUTION:
With  = 0.2107 and  = 17, the probability that there would be AT LEAST 4 orders:
(
≥4
)
=
(
4
)
+
(
5
)
+
(
6
)
+ ⋯ +
(
17
)
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Statistics for Business | Chapter 03: Random Variables
8
=
17
×
(
0.2107
)
×
(
1 − 0.2107
)



With the calculator, you use the following method
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.
)
×
(
− .
)



Note: Press to get sign X in the formula ALPHA + | [ ][ ) ] X
Step 03: Press we can get the result , [ = ] .
d. If the salesperson makes 18 demonstrations, what is the probability that there would be
anywhere from 4 to 8 (both inclusive) orders?
SOLUTION
With  = 0.2107 and  = 18, the probability that there would be FROM 4 TO 8 orders:
(
4 ≤ ≤ 8
)
=
(
4
)
+
(
5
)
+
(
6
)
+
(
7
)
+
(
8
)
=
18
×
(
0.2107
)
×
(
1 − 0.2107
)


With the calculator, you use the following method
Step 01: Press SHIFT + [ log ]
| [ ]
Step 02: Type

( )
×
(
.
)
×
(
− .
)


Note: Press to get sign X in the formula ALPHA + | [ ][ ) ] X
Step 03: Press we can get the result , [ = ] .
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Statistics for Business | Chapter 03: Random Variables
9
e. If the salesperson wants to be at least 90% confident of getting at least 4 orders, at least how
many demonstrations should she make?
SOLUTION:
With  = 0.2107, 
(
≥4
)
≥ 0.9, and unknown , the probability that there would be AT
LEAST 4 orders:
(
≥4
)
=
(
4
)
+
(
5
)
+ ⋯ +
(
)
≥ 0.9
=
×
(
0.2107
)
×
(
1 − 0.2107
)


≥ 0.9
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values
you can guess in the formula until you can find out the value which can satisfy the condition
(
≥
)
≥. .
With
=
17
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.

)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

With
=
20
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.

)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

With
=
25
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.

)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

With
=
30
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.

)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

By using the calculator, we can find out that = 
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Statistics for Business | Chapter 03: Random Variables
10
f. The salesperson has time to make only 22 demonstrations, and she still wants to be at least
90% confident of getting at least 4 orders. She intends to gain this confidence by improving the
quality of her demonstration and thereby improving the chances of getting an order at the end
of a demonstration. At least to what value should this probability be increased in order to gain
the desired con fidence? Your answer should be accurate to four decimal places.
SOLUTION:
With , = 22
(
≥4
)
≥ 0.9, and unknown , the probability that there would be AT LEAST 4
orders:
(
≥4
)
=
(
4
)
+
(
5
)
+ ⋯ +
(
)
≥ 0.9
=
22
×
(
)
×
(
1 −
)



≥ 0.9
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values you
can guess in the formula until you can find out the value which can satisfy the condition
(
≥ ≥
)
. .
With
=
0
.
2107
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.

)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

With
=
0
.
25
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.
)
×
(
−
.

)


Step 03: Press we can get the result , [ = ]
.

With
=
0
.
28
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
×
(
.
)
×
(
−
.

)


Step , 03: Press [ = ] we can get the result
.

By using the calculator, we can find out that ≈0.28
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Statistics for Business | Chapter 03: Random Variables
11
2. Hypergeometric Distribution:  ~ (,,)
Let us derive the formula for when is hypergeometrically distributed. The
(
=
)
number of successes have to come from the successes in the pool, which can happen in
 ways. Th failures have to come from the failures in the pool, which
(
−
) (
−
)
can happen in
−
−
ways. Together the successes and failures can happen
(
−
)
in
−
−
 ways. Finally, there are
 ways of selecting a sample of size .
(
=
)
=

(
,
−
+
)
≤
≤

(
,
)
(
)
=

=
(
/
)
(
)
=

(
−
)
Example | Part 2.2: ( ) Case of Hypergeometric Distribution
PROBLEM
A committee of 7 members is to be formed by selecting members at random from a pool of 14
candidates consisting of 5 women and 9 men.
a. What is the probability that there will be at least three women in the committee?
SOLUTION:
With , (women), and , the probability that there will be at least three = 14 = 5 = 7
women in the committee:
(
3 ≤ ≤ 5
)
=
(
3
)
+
(
4
)
+
(
5
)
=
5
14 − 5
7
−
14
7

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Statistics for Business | Chapter 03: Random Variables
12
With the calculator, you use the following method
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)
(
 −
)
( )
−


Note: Press to get sign X in the formula + ALPHA [ ) ] | [ ]X
Step 03: Press we can get the result , [ = ] .
b. It is desired to increase the chance that there are at least three women in the committee to
80% by doing one of the following:
i. Adding more women to the pool.
SOLUTION
= 7
= + 9 (the pool of  candidates consisting of  women and 9 men)
The probability that there are AT LEAST 3 women in the committee:
(
3 ≤ ≤7
)
=
(
3
)
+
(
4
)
+
(
5
)
+
(
6
)
+
(
7
)
≈ 0.8
=
(
+ 9
)
−
7
−
+ 9
7

=
9
7
−
+ 9
7

≈ 0.8
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values you
can guess in the formula until you can find out the value which can satisfy the condition
(
≤≤
)
≈..
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Statistics for Business | Chapter 03: Random Variables
13
With
=
7
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.

With
=
8
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.

With
=
9
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.

By using the calculator, we can find out . It means that it is desired to increase the chance = 9
that there are at least three women in the committee to 80% by adding 4 more women to the
pool with a pool of 18 candidates consisting of 9 women and 9 men.
ii. Removing some men from the pool.
SOLUTION:
= 7
= 5 + (the pool of  candidates consisting of 5 women and men)
The probability that there are AT LEAST 3 women in the committee:
(
3 ≤ ≤5
)
=
(
3
)
+
(
4
)
+
(
5
)
≈ 0.8
=
5
(
5 +
)
− 5
7
−
5 +
7

=
5
7
−
5 +
7

≈ 0.8
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Statistics for Business | Chapter 03: Random Variables
14
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values
you can guess in the formula until you can find out the value which can satisfy the condition
(
≤≤
)
≈..
With
=
9
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.
With
=
7
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.

With
=
6
Step 01: Press + SHIFT [ log ]
| [ ]
Step 02: Type
(

)

(
−
)
(
+
)

Step 03: Press we can get the result , [ = ]
.

By using the calculator, we can find out . It means that it is desired to increase the = 6
chance that there are at least three women in the committee to 80% by removing 3 men from
the pool with a pool of 11 candidates consisting of 5 women and 6 men.
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International University IU
STATISTICS FOR BUSINESS [IUBA] CHAPTER 03 RANDOM VARIABLES PART 1
1. Random Variable X – Probability P(X) – Cumulative Distribution Function F(X) Random Variable Probability
Cumulative Distribution Function X P(X) F(X) 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.2 0.8 4 0.1 0.9 5 0.1 1.0 1.0 P( x) ≥ 0 for all values x F( x) = P( X ≤ x) = P( x) ables ari P(x) = 1 V dom 03: Ran
2. Expected Value, Variance, and Standard Deviation of Random Variable X hapter
Expected Value (Mean) µ = E( X) = ∑ xP( x) Variance
σ = V( x) = E[(x − µ) ] = ∑ ( x − µ) P( x) siness | C Bu r Standard Deviation σ = SD (X) = V( X) tatistics fo S
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CALCULATOR INSTRUCTIONS FOR STATISTICS
Note: This page is only relevant for CASIO scientific calculator FX-570ES
Computing Mean and Standard Deviation of Random Variable X.
(Chapter 03 | Random Variables)
Step 01: Press MODE 3: STAT 1: 1 - VAR SHIFT + MODE | [SET ] UP REPLAY 4: STA T 1: ON (Frequency ?)
Step 02: Input the data into two columns
Step 03: Compute the expected value of the random variable X
No.01: Put the cursor in the (n+1)th line and FREQ column No.02: Press SHIFT + 1 | [STAT] 5: VAR ables ari 2: V [ = ] dom
Step 04: Compute the standard deviation of the random variable X 03: Ran
No.01: Return the table of data and put the cursor in the (n+1)th line and FREQ column. hapter
No.02: Press [ DEL ] to delete the value of the expected value in step 03.
(This step is very important for you to remember. If you do not delete siness | C
the expected value you had already computed, you will get the wrong Bu r result in this computation) tatistics fo S
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No.03: Press SHIFT + 1 | [STAT] 5: VAR 3: [ = ]
Step 04: Compute the variance of the random variable X
We can get the value of the variance of the random variable X by using the following formula: ( ) = [ ( ) ] Example | Part 1 PROBLEM:
The number of defects in a machine-made product is a random variable X with the
following probability distribution: x P(x) 0 0.1 1 0.2 2 0.3 3 0.3 4 0.1 ables
a. Show that P(x) is a probability distribution. ari
b. Find the probability (1 < ≤3) . V
c. Find the probability (1 < ≤4)
d. Find the cumulative distribution function F(x). dom
e. Find the expected value, variance, and standard deviation of the random variable X. 03: Ran SOLUTION: a. hapter
P( x) = P (0) + P (1) + P( 2 ) + P (3) + P (4)
= 0.1 + 0.2 + 0.3 + 0.3 + 0.1 = 1.0 siness | C
Hence, P(X) is a probability distribution Bu r b. P(1 <
≤ 3 ) = P ( 2) + P( 3 ) = 0.3 + 0.3 = 0.6 tatistics fo S
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International University IU c. P( 1 <
≤ 4) = P( 2) + P( 3) + P( 4) = 0.3 + 0.3 + 0.1 = 0 .7 d. x P(x) F(X 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.3 0.9 4 0.1 1.0 e. General Method ( ) ( ) ( − ) ( ) 0 0.1 0.0 0.441 1 0.2 0.2 0.242 2 0.3 0.6 0.003 3 0.3 0.9 0.243 4 0.1 0.4 0.361 2.1 1.29 Expected Value (Mean): E( X) = xP( x) = 2.1 Variance: V( X) = ( x − µ) P(x) = 1.29 ables ari V Standard Deviation: dom SD( X) = V( X) = √1.29 ≈ 1.1358 03: Ran Calculator Method
To compute the expected value, standard deviation, and variance by
using the calculator, we follow these steps: hapter Step 01: Press MODE 3: STAT 1: 1 - VAR SHIFT + MODE | [SET UP] siness | C REPLAY 4: STAT Bu r 1: ON (Frequency ?)
Step 02: Input the data into two columns tatistics fo S
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Step 03: Compute the expected value of the random variable X
No.01: Put the cursor in the 6th line and FREQ column
No.02: Press SHIFT + 1 | [STAT] 5: VAR 2: [ = ]
(The available result you can get is 2.1)
Step 04: Compute the standard deviation of the random variable X No.01:
Return the table of data and put the cursor
in the 6th line and FREQ column. No.02:
Press [ DEL ] to delete the value of the expected value in step 03.
(This step is very important for you to
remember. If you do not delete the expected
value you had already computed, you will get
the wrong result in this computation) No.03: Press SHIFT + 1 | [STAT] 5: VAR 3: [ = ]
(The available result you can get is 1.13578166
Step 04: Compute the variance of the random variable X Since ( ) = ( ) or ( ) = [ ( ) ] , we can get the result as following ( ) = [ ( ) ] = (1.135781669) = . ables ari V dom 03: Ran hapter siness | C Bu r tatistics fo S
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International University IU PART 2
1. Binomial Distribution: ~ ( , ) Let be the number of trials
be the probability of success
Conditions for a Binominal Random Variable
The trials must be Bernoulli trials in that the outcomes can only be either success or failure.
The outcomes of the trials must be independent.
The probability of success in each trial . ( = ) = ( − ) ( ) = , , …, ( ) = ( ) = ( − )
Example | Part 2.1: (Case of Binomial Distribution) PROBLEM: ables ari V
A salesperson goes door-to-door in a residential area to demonstrate the use of a new
household appliance to potential customers. At the end of a demonstration, the probability that dom
the potential customer would place an order for the product is a constant 0.2107. To perform
satisfactorily on the job, the salesperson needs at least four orders. Assume that each 03: Ran
demonstration is a Bernoulli trial.
a. If the salesperson makes 15 demonstrations, what is the probability that there would be hapter exactly 4 orders? SOLUTION siness | C Bu
With = 0.2107 and = 15, the probability that there would be EXACTLY 4 orders: r ( 15 = 4) =
× ( 0.2107) × ( 1 − 0.2107) 4 tatistics fo S
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With the calculator, you type as following ( ) × ( . ) × ( − . )
After pressing [ = ], we can get the result .
b. If the salesperson makes 16 demonstrations, what is the probability that there would be at most 4 orders? SOLUTION
With = 0.2107 and = 16, the probability that there would be AT MOST 4 orders: ( ≤4) = ( 0) + ( 1) + ( 2) + ( 3) + ( 4) 16 =
× ( 0.2107) × ( 1 − 0.2107)
With the calculator, you use the following method Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ables ( ) × ( . ) × ( − . ) ari V Note: Press ALPHA +
[ ) ] | [ X ] to get sign X in the formula dom
Step 03: Press [ = ], we can get the result . 03: Ran hapter
c. If the salesperson makes 17 demonstrations, what is the probability that there would be at least 4 orders? SOLUTION: siness | C Bu r
With = 0.2107 and = 17, the probability that there would be AT LEAST 4 orders: ( ≥4) = ( 4) + ( 5) + ( 6) + ⋯ + ( 17) tatistics fo S
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International University IU 17 =
× ( 0.2107) × ( 1 − 0.2107)
With the calculator, you use the following method Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) × ( . ) × ( − . ) Note: Press ALPHA + | [ [ ) ] X ] to get sign X in the formula
Step 03: Press [ = ], we can get the result .
d. If the salesperson makes 18 demonstrations, what is the probability that there would be
anywhere from 4 to 8 (both inclusive) orders? SOLUTION
With = 0.2107 and = 18, the probability that there would be FROM 4 TO 8 orders: ( 4 ≤ ≤ 8) = ( 4) + ( 5) + ( 6) + ( 7) + ( 8) ables ari V 18 =
× ( 0.2107) × ( 1 − 0.2107) dom
With the calculator, you use the following method 03: Ran Step 01: Press SHIFT + [ log ] | [ ] hapter Step 02: Type ( ) × ( . ) × ( − . ) siness | C Bu r Note: Press ALPHA + | [ [ ) ] X ] to get sign X in the formula
Step 03: Press [ = ] we can get the result , . tatistics fo S
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e. If the salesperson wants to be at least 90% confident of getting at least 4 orders, at least how
many demonstrations should she make? SOLUTION: With = 0.2107, (
≥4) ≥ 0.9, and unknown , the probability that there would be AT LEAST 4 orders: ( ≥4) = ( 4) + ( 5) + ⋯ + ( ) ≥ 0.9 =
× ( 0.2107) × ( 1 − 0.2107) ≥ 0.9
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values
you can guess in the formula until you can find out the value which can satisfy the condition ( ≥ ) ≥ . . With = 17 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) × ( . ) × ( − . )
Step 03: Press [ = ] we can get the result , . With = 20 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ables ari V ( ) × ( . ) × ( − . ) dom
Step 03: Press [ = ] we can get the result , . With = 25 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type 03: Ran ( ) × ( . ) × ( − . ) hapter
Step 03: Press [ = ] we can get the result , . With = 30 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type siness | C Bu r ( ) × ( . ) × ( − . )
Step 03: Press [ = ] we can get the result , .
By using the calculator, we can find out that = tatistics fo S
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International University IU
f. The salesperson has time to make only 22 demonstrations, and she still wants to be at least
90% confident of getting at least 4 orders. She intends to gain this confidence by improving the
quality of her demonstration and thereby improving the chances of getting an order at the end
of a demonstration. At least to what value should this probability be increased in order to gain
the desired confidence? Your answer should be accurate to four decimal places. SOLUTION:
With = 22, ( ≥4) ≥ 0.9, and unknown , the probability that there would be AT LEAST 4 orders: ( ≥4) = ( 4) + ( 5) + ⋯ + ( ) ≥ 0.9 22 = × ( ) × ( 1 − ) ≥ 0.9
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values you
can guess in the formula until you can find out the value which can satisfy the condition ( ≥ ) ≥ . . With = 0 .2107 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) × ( . ) × ( − . ) ables ari V
Step 03: Press [ = ] we can get the result , . With = 0 .25 Step 01: Press SHIFT + [ log ] | [ ] dom Step 02: Type ( ) × ( . ) × ( − . ) 03: Ran
Step 03: Press [ = ] we can get the result , . hapter With = 0 .28 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) × ( . ) × ( − . ) siness | C Bu r
Step 03: Press [ = ], we can get the result .
By using the calculator, we can find out that ≈0.28 tatistics fo S
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International University IU
2. Hypergeometric Distribution: ~ ( , , )
Let us derive the formula for ( =
) when is hypergeometrically distributed. The
number of successes have to come from the successes in the pool, which can happen in ways. Th(
− ) failures have to come from the ( − ) failures in the pool, which can happen in −
ways. Together the successes and ( − ) failures can happen − in − ways. Finally, there are
ways of selecting a sample of size . − ( = ) = ( , − + ) ≤ ≤ ( , ) ( ) = = ( / ) ( ) = ( − )
Example | Part 2.2: (Case of Hypergeometric Distribution) PROBLEM ables
A committee of 7 members is to be formed by selecting members at random from a pool of 14 ari
candidates consisting of 5 women and 9 men. V dom
a. What is the probability that there will be at least three women in the committee? SOLUTION: 03: Ran With = 14, = 5 (women), and
= 7, the probability that there will be at least three women in the committee: hapter ( 3 ≤ ≤ 5) = ( 3) + ( 4) + ( 5) 5 14 − 5 siness | C 7 − Bu = r 14 7 tatistics fo S
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International University IU
With the calculator, you use the following method Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( − ) Note: Press ALPHA +
[ ) ] | [ X ] to get sign X in the formula
Step 03: Press [ = ] we can get the result , .
b. It is desired to increase the chance that there are at least three women in the committee to
80% by doing one of the following:
i. Adding more women to the pool. SOLUTION = 7 =
+ 9 (the pool of candidates consisting of women and 9 men)
The probability that there are AT LEAST 3 women in the committee: ables ari ( 3 ≤ ≤7) = ( 3) + ( 4) + ( 5) + ( 6) + ( 7) ≈ 0.8 V ( + 9) − 9 dom 7 − = = 7 − ≈ + 9 + 9 0.8 7 7 03: Ran hapter
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values you siness | C
can guess in the formula until you can find out the value which can satisfy the condition ( ≤ ≤ Bu ) ≈ . . r tatistics fo S
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International University IU With = 7 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , . With = 8 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , . With = 9 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , .
By using the calculator, we can find out
= 9. It means that it is desired to increase the chance
that there are at least three women in the committee to 80% by adding 4 more women to the
pool with a pool of 18 candidates consisting of 9 women and 9 men. ables
ii. Removing some men from the pool. ari V SOLUTION: dom = 7 = 5 +
(the pool of candidates consisting of 5 women and men) 03: Ran
The probability that there are AT LEAST 3 women in the committee: hapter ( 3 ≤ ≤5) = ( 3) + ( 4) + ( 5) ≈ 0.8 5 ( 5 + ) − 5 5 7 − siness | C = = 7 − ≈ 0.8 5 + 5 + Bu r 7 7 tatistics fo S
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International University IU
With the calculator, you use the following method
Note: The calculator cannot provide you the final value , so you have to try the different values
you can guess in the formula until you can find out the value which can satisfy the condition ( ≤ ≤ ) ≈ . . With = 9 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , . With = 7 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , . With = 6 Step 01: Press SHIFT + [ log ] | [ ] Step 02: Type ( ) ( − ) ( + )
Step 03: Press [ = ] we can get the result , . ables ari
By using the calculator, we can find out
= 6. It means that it is desired to increase the V
chance that there are at least three women in the committee to 80% by removing 3 men from dom
the pool with a pool of 11 candidates consisting of 5 women and 6 men. 03: Ran hapter siness | C Bu r tatistics fo S
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