April 27 - Kenston School District

HONORS STATISTICS 

Mrs. Garrett Period 2 

Friday April 27, 2012 

1

Daily Agenda 

1. Welcome to class 

2. Please find folder and take 

your seat. 

3. Review simulations 

4. Review homework OTL C6#3 

5. Finish Notes and practice on 6.2 

6. Collect folders 

2

PRACTICE SIMULATION 2 

At a local high school, 95 students have permission to park on campus. Each month the student council holds a 

“golden ticket parking lottery” at a school assembly. The two lucky winners are given reserved parking spots next to 

the school’s main entrance. Last month, the winning tickets were drawn by a student council member from the 

Honors Statistics class. When both golden tickets went to members of that same class, some people thought the 

lottery had been rigged. There are 28 students in the Honors Statistics class, all of whom are eligible to park on 

campus. Design and carry out a simulation to decide whether it’s plausible that the lottery was carried out fairly. 

Follow the 5 steps: 

3

STEP 1: State the problem 

Simulate a golden ticket lottery where there are 95 

eligible students and only 2 tickets are winner. 

Determine the probability that the two winners are from 

the same math class of 28 students. 

STEP 2: State assumptions 

Assume that the tickets are drawn independently. 

All tickets are separated and well mixed. 

4

STEP 3: Assign digits 

LABEL: use two digit numbers between 01 and 95 

Let 01 to 28 represent the students 

in the honors statistics class. 

Let 29 to 95 represent the other eligible students. 

TABLE: use line 139 and pick off 2 digits at a time 

STOPPING 

PROCEDURE: select 2 digit numbers, 

choose 2 different (2 digit) numbers between 01 and 95 

for each simulated drawing 

Do 20 simulations of the golden ticket lottery 

5

STEP 4: Repeat many repetitions ﴾20 simulations of drawing for the golden tickets﴿ 

IDENTIFY SAMPLE: 

Line 139 55588 99404 70708 41098 43563 56934 48394 51719 

Line 140 12975 13258 13048 45144 72321 81940 00360 02428 

Line 141 96767 35964 23822 96012 94591 65194 50842 53372 

DRAW 1 DRAW 2 DRAW 3 DRAW 4 





6

STEP 5: State your conclusion 

7

OTL C6 #3 page 416 and 424 

pg 416: 6.30, 6.32 part c only, 6.36 

pg 424: 6.39, 6.40, 6.41, 6.44 

ASkips are 6.30 and 6.39 

USE PROPER NOTATION !!!! 

P(??) = .?? 

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OTL ANSWERS C6#3 

pg 416: 6.30, 6.32, 6.36 

pg 424: 6.39, 6.40, 6.41, 6.44 

6.30 a) S={0 < x < 24, where x = hours} 

b) S={0 < x < 11,000 where x = number of physicians} 

c) S={0, 1, 2, 3, 4, ..., 12} where x = number of broken eggs 

d) S={0 < x < 200.00, where x = dollars} 

e) S={100 < x < 200 , where x = grams of weight gain} 

6.32 

Toss two coins n = 2 2 = 4 S = { HH, HT, TH, TT } 

Toss three coins n = 2 3 = 8 S = {HHH THH 

HHT THT 

HTH TTH 

Part c) 

HTT TTT 

Toss five coins n = 2 5 = 32 

{HHHHH THHHH 

HHHHT THHHT 

HHHTH THHTH 

HHHTT THHTT 

HHTHH THTHH 

HHTHT THTHT 

HHTTH THTTH 

HHTTT THTTT 

HTHHH TTHHH 

HTHHT TTHHT 

HTHTH TTHTH 

HTHTT TTHTT 

HTTHH TTTHH 

HTTHT TTTHT 

HTTTH TTTTH 

HTTTT TTTTT} 

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Card Facts: 

Use this information for problem 6.36 

There are 52 cards in a deck. 

There are two colors in a deck. 

red and black 

There are four suits in a deck. 

Hearts, Diamonds, 

Spades, and Clubs. 

There are 3 different face cards. 

King, Queen, Jack 

(12 total face cards) 

Numbered cards are 2 thru 10 

6.36 

a) P(red) = 26/52 = .5 

b) P(heart) = 13/52 = .25 

c) P(queen and heart) = 1/52 = .019 

d) P(queen or heart) = 4/52 + 13/52 1/52 = 16/52 = .308 

e) P(queen that is not a heart) = 3/52 = .058 

There are four Aces "A" in a deck. 

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C6#3 homework continued 

6.39 P(CV disease OR cancer) = .45 + .22 = .67 

P(other reason) = 1.67 = .33 

6.40 

a) P(husband does less that his share) = 1 (.12+.61) = 1 .73 = .27 

b) P(husband does at least his share) = .12 + .61 = .73 

6.41 

a) sum = .41+.23+.29+.06+.01= 1 

the probability of all possible outcomes in a sample space = 1 

b) P(not in top 20% of high school class) = 1.41 = .59 

c) P(student in the top 40%) = .41 + .23 = .64 

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6.44 Benford's Law 

The first digits of numbers in legitimate records often follow a 

distribution known as Benford's Law. 

Here is the distribution. 

P(A) = P(first digit is 1) = .301 

P(B) = P(first digit is 6 or greater) = .067+.058+.051+.046 = .222 

P(C) = P(first digit is odd) = .301+.125+.079+.058+.046 = .609 

a) P(D) = P(first digit is less than 4) = .301+.176+.125 = .602 

b) P(B U D) = .222 + .602 = .824 

c) P(D c ) = 1P(D) = 1.602 = .398 

d) P(C ∩ D) = .301 + .125 = .426 

e) P(B ∩ C) = .058+.046 = .104 

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Common probability symbolism 

U = union of events = or 

= intersection of events = and 

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Mutually Exclusive or Disjoint Events 

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Complementary Events 

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INDEPENDENCE IS NOT DISJOINT 

DISJOINT SETS 

could be Independent 

IF two sets are independent their "picture" is intersecting. 

IF a picture of two sets is intersecting the two sets may or 

may not be independent. 

If two sets are independent (set A and set B) then... 

Knowing something about set A does nothing to help me decide if the "item" is in set B 

Potential independent situation .... 

GIRLS 

BOYS 

Male 

Staff member 

18

IF two sets are independent this 

equation must be TRUE 

P(A ∩ B) = P(A)P(B) 

.3 

.4 .2 .1 

does .2 = .6 x .3 

.2 = .18 

NO 

can you change the probabilities 

and make the sets independent? 

so sets are not independent 

Let's try again ... 

.1 

.5 .3 .1 

19

classwork pg 430: 6.46, 6.47, 6.48 

6.46 Defective Chips 

P(all 12 chips work properly) = 1P(not work) = (1.05) = .95 

(but 12 in a row) soooo .95 12 = .54 

6.47 One cannot conclude that (.26)(.16) = .0416 because the 

events are not known to be INDEPENDENT 

(one would think college graduates are less likely to be 

machine operators or laborers) 

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6.48 A = person chosen is male 

B = person chosen is 25 years old or older 

a) Explain why P(A) = .44 7317/16639 = .4397 

b) Find P(B) = (3494+2630)/16639 = 0.36805 

c) Find P(male and at least 25 years old) = P(A ∩ B) 

using the table P(A ∩ B) = (1598+970)/16639 

= 2568/16639 = 0.1543 

Are A and B independent events ? 

Does P(A)xP(B) = P(A ∩ B) ? 

0.44 x 0.368 = 0.1543 NO 

0.16192 does not equal 0.1543 

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C6 #4 due 

Monday 4/30 

pg 430: 6.45, 6.49, 6.50, 6.51 

Read and notes pg 435 452 

A SKIP 6.45 

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class practice exercises 

Pg 432: 6.56, 6.57, 6.58, 6.63, 6.64 

6.56 Show that non-empty independent events must intersect. 

a) if A=non empty and B= non empty 

then independent means that P(A)xP(B)= P(A∩B) > 0 

b) if A∩B is empty, then P(A∩B)=0, which contradicts 

independence 

c) Find an example of two events that are neither disjoint or 

independent. 

6.57 New Census categories 

a) verify the table is legit 

b) P(A) = (.000+.003+.060+.062)=.125 

c) B c = person chosen is NOT white 

P(B c ) = (.000+.003+.062+.036+.121+.027) =.249 

or P(B) = .060+.691=.751 

so P(B c ) = 1-.751 = .249 

d) the person chosen is non-Hispanic White = P(A c nB) 

P(A c ∩B) = .691 (just pick it off the table) 

6.58 Being Hispanic 

are the events A and B independent? 

Does P(A∩B) = P(A)P(B)? 

P(A∩B) = .060 P(A)P(B) = (.125)(.751)=.0939 

.060=.0939 SO NOT INDEPENDENT 

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April 27 - Kenston School District

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