===============================================================================
114
Block 2
Chaires
== Begin 5.1 ================================================================ 1 / 5 / 06
1
A Random Variable x is a rule that matches each outcome of an experiment with a number.
Each number produced by the rule x is called A Value of x.
Page 218
--------------------------------------------------------------------------------------------------------------------------------------Consider the experiment: Toss a coin 3 times.
2  2  2 = 8 outcomes.
Let x be the rule: Match (each outcome) with (the number of heads in the outcome).
The 8 outcomes of the experiment x are in {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
The values of the random variable x are:
3 , 2 , 2 , 1 , 2 , 1 , 1 , 0
===============================================================================
2 The Frequency f of a value is the number of times the value occurs.
1 above has frequency = 3.
===============================================================================
3 A Frequency Distribution shows each different value of x and its frequency. See below: 1st screen.
===============================================================================
f
4 The Relative Frequency of x or The Probability of x is P(x) =
if like values of x are equally likely.
f
===============================================================================
5 A Probability Distribution shows each different value of x and its probability. See below: 4th screen. Pg 220
--------------------------------------------------------------------------------------------------------------------------------------x 12

 1 .5
Form the frequency distribution and the probability distribution for 3, 2, 2, 1, 2, 1, 1, 0.  =
N
8

 FREQDIST

f
x
f
The frequency distribution
The frequency histogram
The probability distribution
x
0
1
2
3
sum
f
1
3
3
1
8
P(x)
1/8 = .125
3/8 = .375
3/8 = .375
1/8 = .125
1
The probability distribution in L1, L2

P(x) =
f
.
f
P(3) =
P(0) =
1
= .125
8
1
= .125.
8
Mean =
 (x  f)
N
=
0363
= 1.5
8
N=f = 8
Frequency distribution formula
===============================================================================
A Set is Countable if it is a finite set or its elements can be paired, one-to-one, with the numbers in {1,2,3,4,…}.
A Set is Uncountable if it is not countable.
--------------------------------------------------------------------------------------------------------------------------------------{3,5,8,2,10,1} is a countable because it is finite. {1, 3, 5, 7,…} is countable, even though it is infinite, because
its elements can be paired, one-to-one, with the numbers in {1, 2, 3, 4,…}.
The set of numbers between 0 and 1 is uncountable. Its elements cannot be paired with those in {1,2,3,4,…}.
For any uncountable set, between two of its elements there is another element.
1
0
1
===============================================================================
===============================================================================
6 A Discrete Random Variable x has a countable number of values.
Its distribution is graphed as a histogram.
Page 218
===============================================================================
7 A Continuous Random Variable x has an uncountable number of values.
Page 219
Its distribution is graphed as a smooth curve.
===============================================================================
8 A Parameter is a number that describes a population.  and  are parameters.
===============================================================================
9 The Mean  or The Expected Value E(x) of a discrete random variable x is  = E(x) =  x  P(x)  . Pg 223
===============================================================================
The Standard Deviation  of a discrete random variable x is  =
  (x  )2  P(x) 
Page 223
--------------------------------------------------------------------------------------------------------------------------------------Page 223, 2

x
 PROBDIST
P(x)

.27 .31
.18 .09 .15
Med is not on the 82
The decimals are not on graph
The probability distribution
    P(X  3) = .76
The 83 screen
P(X  3) = P(1) + P(2) + P(3) = .27+.31+.18 = .76
--------------------------------------------------------------------------------------------------------------------------------------The hand computed method. For demonstration only.
x
1
2
3
4
5
P(x)
.27
.31
.18
.09
.15
x P(x)
.27
.62
.54
.36
.75
P(x) = 1
xP(x) =2.54
 = mean
x–
-1.54
-.54
.46
1.46
2.46
Page 223
(x – )2
2.3716
.2916
.2116
2.132
6.0516
(x – )2  P(x)
.640332
.090396
.038088
.19188
.90774
((x-)2 P(x)) = 1.868436
2 = variance
 = 1.868436  1.3669
===============================================================================
2
== Page 230, 12 ===================================================================
You buy 6 chances, at $5 each, to win $2000. Find the expected earnings, and expected gain. Computed below.
G
Gain if you win = 2000 – 30 = 1970.
1970
-30
Gain if you lose = -30.
Sums
P(G)
6
2852
2846
2852
1
G  P(G)
4.1445
Probability of winning.
-29.9369
Probability of losing.
-25.7924
Expected gain.
Assume that the prize was not donated to the club.
$25.7924 is profit to The Center .
The $30 paid to play this game is split as follows: 
$4.2076 goes to the prize fund of $2000.
The $4.2076 is called the expected earnings, also found as follows: Winnings  P(winning) = 2000 
6
2852
The Center takes in $5  2852 = $14,260. $14,260 – $2000 = $12260 profit to The Center.
--------------------------------------------------------------------------------------------------------------------------------------6  2852
2846  2852
expected gain
PROBDIST


Remark: In a “fair game” the expected gain is 0. That is, in the long run, you break even.
== Begin 5.2 and 5.3 ===============================================================
10 A Binomial Experiment occurs when: 1. the experiment is repeated n times,
Page 233
2. there are two outcomes, success S or failure F,
3. the probability p, of success, is the same for each trial.
===============================================================================
11 A Binomial Random Variable r has values that are the number of successes in a binomial experiment.
The number of successes will be r: 0, 1, 2, …, n, meaning from 0 (no successes) to n (all successes).
===============================================================================
12 A Binomial Distribution is a probability distribution of the successes and the corresponding probabilities.
--------------------------------------------------------------------------------------------------------------------------------------A basketball player is known to make 70% of her shots. She takes 5 shots. The binomial distribution is:
r
0
1
2
3
4
5
Here n = 5 and p = .70.
P(r) .00243 .02835 .1323 .3087 .36015 .16807
The probability that she makes none of the 5 shots is .00243, written: P(r = 0) = .00243, or P(0)  .2%.
The probability that she makes 4 out of 5 shots is .36015, written: P(r = 4) = .36015, or P(4)  36%.
The probability of making more than 2 out of five, written P(r > 2) = P(r  3)  .8369  84%
3
P(3) + P(4) + P(5)  .84
We will see how to find the binomial probabilities on the next page.
===============================================================================
===============================================================================
A Binomial Probability Distribution is given by P(r) = Cn,r  pr  qn-r , for r = 0, 1, 2, 3,…, n where
n is the number of trials,
p is the probability of success on a single trial,
r is the number of successes,
q is the probability of failure on a single trial.
q = 1 – p.
Cn,r is the number of ways that r successes can occur in n trials.
Page 236
===============================================================================
Optional: To hand compute Cn,r , we need to have the following concept.
===============================================================================
n Factorial , written n! , is defined as follows: For n > 1, n! = 1  2  3 …  n.
0! = 1! = 1.
--------------------------------------------------------------------------------------------------------------------------------------2! = 1  2 = 2.
10! = 1  2  3 …  10 = 3,628,800
Calculator  10   4 
===============================================================================
n
n!
The Binomial Coefficient Cn,r is a number given by Cn,r =
. Other symbols for Cn,r : nCr or  
r! ( n  r )!
r 
--------------------------------------------------------------------------------------------------------------------------------------r
n
 The answer is 10. See below.
In how many ways can 3 successes occur in 5 trials?
C5,3 =
1 2  3  4  5
5!
5!
=
=
= 10.
1

2

3

1

2
3! (5  3)!
3!  2!
Calculator  5   3 3 
Here are the 10 ways to arrange sssff: sssff, ssffs, ssfsf, sffss, sfsfs, sfssf, fsssf, fssfs, fsfss, ffsss.
These are the ways you can have 3 successes and 2 failures.
===============================================================================
Page 238, 6 This binomial experiment has n = 10 and p = .70. If a seed germinates, then this is success.
Let X be the number of successes. (a) Find P(X = 8).
(b) Find P(X  8) = P(8) + P(9) + P(10)
 BINOMIAL
P(X = 8)  .233.
P(X  8)  .383.
----------------------------------------------------------------------------------------------------------------------------- -------------------------------------
To compute P(X = 8) by formula: P(r) = Cn,r  pr  qn-r
10C8

10   3 8 
P(8) = C10,8  .708  .302 = 45  .708  .302  .23347…
45  .70  8  .30  2 
 45
There are 45 ways to arrange ssss ssss ff.
===============================================================================
13 The Mean  of a Binomial Distribution is  = np.
 is also called The Expected Value
Page 251
===============================================================================
14 The Standard Deviation  of a Binomial Distribution is  = npq = q .
Page 251
===============================================================================
4
===============================================================================
Page 249, 6 This binomial experiment has n = 6 and p = .50. Make the binomial distribution and histogram.
Page 251, 7 (a) Find the expected value. (b) Find the standard deviation .
 BINOMIAL

Successes
 QUIT
0 1 2 3 4 5 6
 = 3,   1.22
histogram
The binomial distribution
By formula:  = np = 6  .50 = 3.
=
npq =
6  .50  .50 = 1.2247…
 ( 10 .50.50 ) 
===============================================================================
Summary: Discrete random variables. Look at the numbers generated by tossing 3 coins and counting heads from sheet 1.
Use  DESCRIPT to find  = 1.5 and  = .866….
A) Put the raw data : 3, 2, 2, 1, 2, 1, 1, 0, into L1.
x
M
B) Put the frequency distribution:
x
f
L1
L2
0
1
2
3
1
3
3
1
into L1, L2. Use  FREQDIST to find  = 1.5 and  = .866….
M
S
C) This is a binomial distribution with n = 3 and p = .5. Use  BINOMIAL to find  = 1.5 and  = .866….
M
S
Using formulas:
A)  =
x 12
=
= 1.5.
N
8
( x   ) 2
(3  1.5)2  (2  1.5)2  (2  1.5)2  (1  1.5)2  (2  1.5)2  (1  1.5)2  (1  1.5)2  (0  1.5)2
=
=
N
8
 .866…
Probability Distribution
1
3
3
1
B)  =  x  P(x)  = 0 
+ 1
+ 2
+ 3  = 1.5
8
8
8
8
=
( x  )
2
1
3
3
1
 P( x ) = (0  1.5)2   (1  1.5)2   (2  1.5)2   (3  1.5)2   .866….
8
8
8
8
C)  = np = 3  .5 = 1.5
This formula is on sheet 2, below note 9.
5
x
P(x)
L1
L2
0
1
2
3
1/8
3/8
3/8
1/8
 = npq = 3  .5  .5  .866….
===============================================================================
=== End 5.3 = Start 6.1 =============================================================
2 2
A Normal Probability Distribution is a continuous distribution given by f(x) = 1  (2).5  e.5( x )  ,
where   3.142 and e  2.718. Its graph is called a normal curve, sometimes called a “bell curve”.
===============================================================================
A Normal Curve has these properties:
Page 293
1. There is a vertical axis of symmetry through .
2. The x-axis is an asymptote.
3. There are inflection points at  –  and  +.
4. The area between the curve and the x-axis = 1 = 100%. This area represents the entire distribution.
5. About 68.3% of the data is within 1 of .
About 95.5% of the data is within 2 of .
The percents are approximate.
About 99.7% of the data is within 3 of .
99.7%
95%
2.3%
2.3%
16%
.0013
.
0 13%
68%
14%
34%
16%
2%
X
 - 3
 - 2
-

+
 + 2
 + 3
===============================================================================
1 The Graph of a Normal Distribution is a normal curve.
===============================================================================
2 The Empirical Rule states: For a distribution that is mound shaped,
Pg 296
a) about 68% of the data are within 1 of ,
b) about 95% of the data are within 2 of ,
c) about 100% of the data are within 3 of .
===============================================================================
Page 296,1 The number of hours x, of life of radios, is a normal distribution with  = 600 and  = 100.
If one radio is selected, find P(600 < x < 700).
34%
300
400
500
-
600

P(x > 800)  .02
34%
700
 +
800
900 x
By the empirical rule, 68% of the data is within 1 of . Because of symmetry, 34% of the data is in each
region shown above. So P(600 < x < 700)  34%.
There is a 34% chance that a radio will last between 600 and 700 hours.
 NORMAL
6
x
===============================================================================
===============================================================================
3 A Control Chart is a graph of data, over a period of time, comparing the actual distribution with Page 298
the expected distribution.
===============================================================================
A random variable is In Control if it remains, more or less, within 2 of  as it is observed over time.
===============================================================================
4 Out of Control Warning Signals: A random variable is probably Out of Control if:
Page 301
I one value is more than 3 from .
P(I)  .003.
II nine consecutive values are on the same side of .
P(II)  .004.
III 2 of 3 consecutive values are beyond 2 on the same side of . P(III)  .004.
===============================================================================
Page 301, 12b The visitors treated have  = 21.7 and  = 4.2 Make a control chart. Find warning signals.
Day
# Treated
1
20
2
15
3
12
Put all data in L1 and L2
4
21
5
24
6
28
7
32
8
36
9
35
10
37
Put in L1
Put in L2
 CONCHART

+3
+2
34.4
30.1

21.7
The standard deviations intervals
25.9
To see them press    etc
 INTERVAL
17.5
 -2
 -3
13.3
9.1
1 2
3 4 5 6 7 8 9 10
Days 8, 9,10, each give Warning Signal I.
days
Days 6, 7, 8 together give Warning Signal III.
== Begin 6.2 =====================================================================
x 
5 The z-score of x is z =
. z is the number of standard deviations that x is from .
Page 309

The z-score is also called the z-value or standard value of x.
x is a value in a distribution.
===============================================================================
6 The raw score x of z is x = z + .
This formula computes the value of x that corresponds to the value of z.
-------------------------------------------------------------------------------------------------------------------------------------
7
To derive the formula in 6, solve formula 5 for x: z =
x 

z =x– 

z+=x 
x = z + . Page 311
===============================================================================
===============================================================================
Page 310, 4 A normal distribution has  = 8 and  = .5. A value of x is 6.9. Find the z-score of 6.9.
 ZTXVALUE
(Use n=1 because 6.9 is a single number from the population.)
Optional: By formula: z =
The z score of 6.9 = -2.2
x 
(6.9  8)
=
= -2.2

.5
--------------------------------------------------------------------------------------------------------------------------------------Graph the normal distribution showing the x and z standard deviation intervals. Show 6.9 on the graph.
2.3 %.
2.3 %
2.5%
2.5%
95%
6.5
-3
7
7.5
8
8.5
9
-2
-1
0
1
2
9.5 x values
3
z values
6.9 is 2.2 below the mean 8.
===============================================================================
Page 311, 5 A value x in the above distribution has a z-score of –3. Find its raw score.
Formula:
 ZTXVALUE
x = z + 
= -3  .5 + 8.
= 6.5
The x value for z = –3 is 6.5
(Use n=1 because we have a single number from the population.)
6.5 is 3 below the mean. See graph above.
===============================================================================
7 The STANDARD Normal Distribution , SND, is a normal distribution with  = 0 and  = 1.
Its graph is the standard normal curve, SNC, also called the z curve.
Page 312
--------------------------------------------------------------------------------------------------------------------------------------Any pizza with less than 6.5 oz. of cheese is more than 3 below the mean 8.


.5
The standard normal curve is the graph of the equation f(x) = 2e x
-------------------------------------------------------------------------------------------------------------------------------------8 The area under the entire curve = 1. 1 = 100%
The area to the right of 0 = .5.
.5 = .5000 = 50%
100%
2
50% 50%
8
-3
-2
-1
0
1
2
3
z
z
An area under a curve corresponds to the probability of selecting a number in that area. P(z > 0) = .5
===============================================================================
===============================================================================
Areas Under A Normal Curve
-------------------------------------------------------------------------------------------------------------------------------------- NORMAL finds areas under a normal curve. It finds Left, Right, and Between areas.
--------------------------------------------------------------------------------------------------------------------------------------Page 314, 6b In the SNC, find (the area to the left of 1.18) = P(z < 1.18).
1: A Left Area
 NORMAL

Z
P(z < 1.18)  .881 = 88.1%
1E99 is the TI symbol for infinity.
--------------------------------------------------------------------------------------------------------------------------------------Page 318, 9a In the SNC, find P(z  1.15). P(z >1.15) = (the area to the right of 1.15).
2: A Right Area
 NORMAL

Z
P(z > 1.15)  .1251 = 12.51%.
In a continuous distribution
P(z  1.15) = P(z > 1.15),
because the boundary at 1.15 has no thickness.
--------------------------------------------------------------------------------------------------------------------------------------Page 318b, In the SNC, find (the area from -1.78 to .35) = P(-1.78 < z < .35).
3: A Between Area
 NORMAL

Z
P(-1.78 < z <.35)  .5993= 59.93%.
--------------------------------------------------------------------------------------------------------------------------------------Draw the area of interest on paper before using the calculator.
Small tail areas may not be seen on the graph screen.
Graph P(z < -2.7)
9
Note: Areas found with PRGM NORMAL are exact, whereas areas found using the table are approximate.
The areas in the table are rounded to 4 places. Also, the z values used are often rounded numbers.
So your results may be slightly different from those given in our book.
===============================================================================
===============================================================================
How to use the table to find areas under the SNC (Optional)
===============================================================================
 NORMAL
A Left Area is the area to the left of the number z1.
1 1 1
To find the area, look up z1. P(z < 1.44) = .9251
R = 1.44
-3
-2
-1
0
1
2
3
z
z1 = 1.44
R
===============================================================================
A Right Area is the area to the right of z1. To find it, look up z1 and subtract from 1.
1 – .9911 = .0089
.0089 = P(z > 2.37)
-3
-2
-1
0
1
2
 NORMAL
1 1 2
L = 2.37
z
3
2.37 = z1
===============================================================================
 NORMAL
A Between Area is the area between two numbers z1 and z2.
1 1 3
.0572 To find it, look up z2 and z1, then subtract.
L = 1.48
R = 2.25
z
.9878 – .9306 = .0572
-3
-2
-1
0
1
2
3
z1 = 1.48
2.25 = z2
L
R
===============================================================================
A Two Tail Area is the sum of two tails. To find this area, find the left tail and double it.
P(z < -2.5 or z > 2.5) = .0124
2  P(z < -2.5)
-3
-2
-1
0
1
10
2
3
z
2(.0062) = .0124
z1 = -2.5
2.5 = z2
===============================================================================
Summary
To find a Left Area, look up z1.
To find a Right Area, look up z1 and subtract from 1.
To find a Between Area, look up z2 and z1 and subtract.
===============================================================================
==================================================================
Note: In a normal, or any continuous, distribution the area found using an open inequality, such as <,
is the same as the area found using a closed inequality, such as .
For example: P(z < 1.44) = P(z  1.44). This is true because the boundary line at 1.44 has no thickness.
Including 1.44 in the area adds nothing to the area and not including 1.44 takes nothing from the area.
However, this is not true for discrete distributions, such as the binomial distribution, where P(r < 5)  P(r  5).
Here the 5 bar has thickness and including it adds to the area.
P(r < 5) = P(3) + P(4).
P(r  5) = P(3) + P(4) + P(5).
3
4 5
6
7
=== End Block 2 ==================================================================
11
114
Language Test Statements
Erdeky
--------------------------------------------------------------------------------------------------------------------------------------1 Define:
A Random Variable x
2 Define:
The Frequency f
3 Define: A Frequency Distribution
4 Define: The Relative Frequency of x
5 Define:
A Probability Distribution
6 Define: A Discrete Random Variable x
7 Define: A Continuous Random Variable x
8 Define: A Parameter
9 Complete: The Mean  or The Expected Value E(x) of a discrete random variable x is  = E(x) =
10 Complete: A Binomial Experiment occurs when:
1
2
3
11 Define: A Binomial Random Variable r
12 Define: A Binomial Distribution
13 Complete: The Mean  of a Binomial Distribution or Expected Value is  =
14 Complete: The Standard Deviation  of a Binomial Distribution is  =
1 Complete: The Graph of a Normal Distribution is
2 Complete: The Empirical Rule states: For a distribution that is ___________________________,
a) about _______% of the data are within _________________,
b) about _______% of the data are within _________________,
c) about _______% of the data are within _________________.
3 Define: A Control Chart
4 Complete: Out of Control Warning Signals
I
A random variable is probably out of control if:
II
III.
5 Complete: The z-score of x , a value in a distribution, is z =__________________________
z is the number of __________________ that x is from________.
12
6 Complete: The raw score x of z is x =
7 Complete: The Standard Normal Distribution , SND, is a normal distribution with ___________________
8 Complete: The area under the entire curve = ___________. The area to the right of 0 = _____________
===============================================================================
114 B2 Problems
5.1 ---------------------------------------------------------------------------------------------------------------------------------227, 2 a) C
b) C
c) D
d) C
e) D
228, 5

PROBDIST
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
228, 7
L1
36
37
38
39
40
41
42
43
44
45
229, 9 
L2
6
10
11
20
26
32
34
28
25
16
L3
FREQDIST
(c) P(39,40,41,42,43) =
PROBDIST
d
b
e
20  26  32  34  28
=.673  67%
208
L1
36
37
38
39
40
41
42
43
44
45
L2
.02885
.04808
.05288
.09615
.125
.15385
.16346
.13462
.12019
.07692
L3
6
10
11
20
26
32
34
28
25
16
 QUIT  
a

c
--------------------------------------------------------------------------------------------------------------------------------------L2
L3
230, 11 L1
35 – 15
15 / 719
PROBDIST
Expected Gain. This is her
-15
704 / 719
contribution to the club.
15 – 14.2698 = .73 is her
P(losing)
P(winning)
$.73 is her expected winnings.
The club uses the $15 as follows: $0.73 is to pay for the dinner. $14.27 is for the club’s new equipment.
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230, 13
PROBDIST
P(not dying at age 60 – 64 )
13
(b) The expected loss on Jim when he is age 60 to 64 = $3497.50.
(c) To make an expected profit of $700, the company must charge 700 + 3497.50 = $4197.50.
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5.2 ---------------------------------------------------------------------------------------------------------------------------------242, 1 Trial: flip a coin. Success x: heads n = 3, p = .5
BINOMIAL
(a) P(3H) = .125
(b) P(2H) = .375
(c)
P(x  2) = .5
(d) P(No H) = .125
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242, 3a Trial: Would you marry the same woman again. Success x: yes.
n = 10,
Find P(x  7)
Also, find P(x less than half) = P(x < 5) = P(x  4)
BINOMIAL
p = .80
P(x  7)  88%
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244, 11d
n = 16,
P(L) = .95.  P(x > 14) = P(x  15) = .81
BINOMIAL
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245, 15d
n = 10, P(never develop an eye disease) = .70  P(x  6)  85%
BINOMIAL
5.3 -------------------------------------------------------------------------------------------------------------------------------255, 2a
BINOMIAL
256, 3a
BINOMIAL

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14
257, 5
Trial: look for a friend. Success: find a friend.
BINOMIAL
n = 6,
p =.70

Mean = 4.2 Standard Deviation = 1.12. The Service should find about 4.2 (about 4) friends.
--------------------------------------------------------------------------------------------------------------------------------------Ch 5 Review ---------------------------------------------------------------------------------------------------------------------282, 7b BINOMIAL
6.1 ---------------------------------------------------------------------------------------------------------------------------------305, 7 The Empirical Rule rules.
(
(
(
!
)
)
)
57.5
60
62.5
65
67.5
70
72.5
x = the heights of college women.
(a) P(x > 65) = 50% (b) P(x < 65) = 50% (c) P(62.5 < x < 67.5)  68% (d) P(60 < x < 70)  95%
--------------------------------------------------------------------------------------------------------------------------------------306, 13a Put all 12 months of data (from problem 13a ) in L1 and L2.
 CONCHART
An out-of-control signal is at months 9, 10 ,11.
Also at 10, 11,12.
The trend is not the mean 615.1. Loans are down.
6.2 ---------------------------------------------------------------------------------------------------------------------------------319, 1 a  = 0 in a Z distribution.  R, Ju, L d  ZTXVALUE
Robert
--------------------------------------------------------------------------------------------------------------------------------------320, 5f ZXSCORE
P(z > 2.58 ) = P(x > 5999.6)
--------------------------------------------------------------------------------------------------------------------------------------321, 11 NORMAL
321, 15 NORMAL
15
321, 19 NORMAL
16