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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
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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
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2 The Frequency f of a value is the number of times the value occurs. 1 above has frequency = 3.
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3 A Frequency Distribution shows each different value of x and its frequency. See below: 1st screen.
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4 The Relative Frequency of x or The Probability of x is P(x) = if like values of x are equally likely.
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5 A Probability Distribution shows each different value of x and its probability. See below: 4th screen. Pg 220
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Form the frequency distribution and the probability distribution for 3, 2, 2, 1, 2, 1, 1, 0. =
FREQDIST f
The frequency distribution The frequency histogram The probability distribution in L1, L2
The probability distribution
P(x) = . P(0) = = .125.
P(3) = = .125 Mean = = = 1.5
N=f = 8 Frequency distribution formula
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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.
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{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. 0 1
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6 A Discrete Random Variable x has a countable number of values.
Its distribution is graphed as a histogram. Page 218
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7 A Continuous Random Variable x has an uncountable number of values. Page 219
Its distribution is graphed as a smooth curve.
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8 A Parameter is a number that describes a population. and are parameters.
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9 The Mean or The Expected Value E(x) of a discrete random variable x is = E(x) = .Pg 223
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The Standard Deviation of a discrete random variable x is = Page 223
------Page 223, 2
PROBDIST
The probability distribution P(X 3)= .76
P(X 3) = P(1) + P(2) + P(3) = .27+.31+.18 = .76
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The hand computed method. For demonstration only. Page 223
=1.3669
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== Page 230, 12 ======
You buy 6 chances, at $5 each, to win $2000. Find the expected earnings, and expected gain. Computed below.
Gain if you win = 2000 – 30 = 1970. Probability of winning.
Gain if you lose = -30. Probability of losing.
Expected gain.
The $30 paid to play this game is split as follows:
The $4.2076 is called the expected earnings, also found as follows: Winnings P(winning) =
The Center takes in $5 2852 = $14,260. $14,260 – $2000 = $12260 profit to The Center.
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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 Sor failure F,
3. the probability p, of success, is the same for each trial.
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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).
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12 A Binomial Distribution is a probability distribution of the successes and the corresponding probabilities.
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A basketball player is known to make 70% of her shots. She takes 5 shots. The binomial distribution is:
Here n = 5 and p = .70.
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%
We will see how to find the binomial probabilities on the next page.
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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
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Optional: To hand compute Cn,r , we need to have the following concept.
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n Factorial , written n! , is defined as follows: For n > 1, n! = 1 2 3 … n. 0! = 1! = 1.
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2! = 1 2 = 2. 10! = 1 2 3 … 10 = 3,628,800 Calculator 10 4
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The Binomial Coefficient Cn,r is a number given by Cn,r = . Other symbols for Cn,r: nCr or
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In how many ways can 3 successes occur in 5 trials? The answer is 10. See below.
C5,3 = = = = 10. 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.
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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.
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To compute P(X = 8) by formula: P(r) = Cn,r pr qn-r P(8) = C10,8 .708 .302 = 45 .708 .302 .23347…
10C8 10 3 8 45 45 .70 8 .30 2
There are 45 ways to arrange ssss ssss ff.
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13 The Mean of a Binomial Distribution is = np. is also called The Expected Value Page 251
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14 The Standard Deviation of a Binomial Distribution is = = . Page 251
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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 QUIT
= 3, 1.22 histogram The binomial distribution
By formula: = np = 6 .50 = 3.
= = = 1.2247…
( 10 .50.50 )
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Summary: Discrete random variables. Look at the numbers generated by tossing 3 coins and counting heads from sheet 1.
A) Put the raw data : 3, 2, 2, 1, 2, 1, 1, 0, into L1. Use DESCRIPTto find = 1.5 and = .866….
M x
B) Put the frequency distribution: 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) = = = 1.5.
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.866…
Probability Distribution
B) = = 0 + 1 + 2 + 3 = 1.5
= = .866….
C) = np = 3 .5 = 1.5
= = .866….
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=== End 5.3 = Start 6.1 ======
A Normal Probability Distribution is a continuous distribution given by f(x) = ,
where 3.142 and e 2.718. Itsgraph is called a normal curve, sometimes called a “bell curve”.
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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 .
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1 The Graph of a Normal Distribution is a normal curve.
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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 .
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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). P(x > 800) .02
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
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3 A Control Chart is a graph of data, over a period of time, comparing the actual distribution with Page 298
the expected distribution.
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A random variable is In Control if it remains, more or less, within 2 of as it is observed over time.
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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.
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Page 301, 12b The visitors treated have = 21.7 and = 4.2 Make a control chart. Find warning signals.
Put all data in L1 and L2 CONCHART
The standard deviations intervals
To see them press etc
INTERVAL
Days 8, 9,10, each give Warning Signal I. Days 6, 7, 8 together give Warning Signal III.
== Begin 6.2 ======
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.
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6 The raw score x of z is x = z + . This formula computes the value of x that corresponds to the value of z.
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To derive the formula in 6, solve formula 5 for x: z = z = x – z + = x
x = z + . Page 311
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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 singlenumber from the population.) The z score of 6.9 = -2.2
Optional: By formula: z = = = -2.2
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Graph the normal distribution showing the x and z standard deviation intervals. Show 6.9 on the graph.
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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
(Use n=1 because we have a single number from the population.) The x value for z = –3 is 6.5
Any pizza with less than 6.5 oz. of cheese is more than 3below the mean 8. 6.5 is 3below the mean. See graph above.
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7 The STANDARDNormal 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
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The standard normal curve is the graph of the equation f(x) =
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8 The area under the entire curve = 1. 1 = 100% The area to the right of 0 = .5. .5 = .5000 = 50%
z z
An area under a curve corresponds to the probability of selecting a number in that area. P(z > 0) = .5
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Areas Under A Normal Curve
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NORMAL finds areas under a normal curve. It finds Left, Right, and Between areas.
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Page 314, 6b In the SNC, find (the area to the left of 1.18) = P(z < 1.18). 1: A Left Area
NORMAL
P(z < 1.18) .881 = 88.1%
1E99 is the TI symbol for infinity.
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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
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.
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Page 318b, In the SNC, find (the area from -1.78 to .35) = P(-1.78 < z < .35). 3:A Between Area
NORMAL
P(-1.78 < z <.35) .5993= 59.93%.
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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)
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.
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How to use the table to find areas under the SNC (Optional)
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A Left Area is the area to the left of the number z1.
To find the area, look up z1. P(z < 1.44) = .9251
z
z1 = 1.44
R
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A Right Area is the area to the right of z1. To find it, look up z1andsubtract from 1.
1 – .9911 = .0089
.0089 = P(z > 2.37)
z
2.37 = z1
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A Between Area is the area between two numbers z1 and z2.
.0572 To find it, look up z2and z1, thensubtract.
z .9878 – .9306 = .0572
z1 = 1.48 2.25 = z2
L R
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A Two Tail Area is the sum of two tails. To find this area, find the left tailanddouble it.
P(z < -2.5 or z > 2.5) = .0124
2 P(z <-2.5)
z 2(.0062) = .0124
z1 = -2.5 2.5 = z2
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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.
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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).
=== End Block 2 ======
114 Language Test Statements Erdeky
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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 A random variable is probably out of control if:
I
II
III.
5 Complete: The z-score of x , a value in adistribution, is z =______
z is the number of ______that x is from______.
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 = ______
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114 B2 Problems
5.1------
227, 2 a) C b) C c) D d) C e) D
228, 5 PROBDIST
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228, 7
FREQDIST
(c) P(39,40,41,42,43) = =.673 67%
229, 9 PROBDIST d e a QUIT
b c
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230, 11
PROBDIST Expected Gain. This is her
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 )
(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 (c) P(x 2) = .5
(b) P(2H) = .375 (d) P(No H) = .125
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242, 3a Trial: Would you marry the same woman again. Success x: yes. n = 10, p = .80
Find P(x 7) Also, find P(x less than half) = P(x < 5) = P(x 4)
BINOMIAL
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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257, 5 Trial: look for a friend. Success: find a friend. n = 6, p =.70
BINOMIAL
Mean = 4.2 Standard Deviation = 1.12. The Service should find about 4.2 (about 4) friends.
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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%
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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
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320, 5f ZXSCORE
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321, 11 NORMAL
321, 15 NORMAL
321, 19 NORMAL
1