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Business Statistics

Lecture 7

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Hyper geometric Probability Distribution

• The Probability function

• Where N = no. of elements in population• n = no. of trials• r = no. of successes in the population• x = no. of successes in the trial

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Mean & Variance

• The mean

• The variance

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Example 47

• Suppose N=15 and r = 4. Compute the hyper geometric probability of x=3 and n=10?

• Solution:

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Example 49Blackjack or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards, Face Cards (jacks, queens and kings) and tens have a point value of 10. Aces have a point value of 1 or 11. A 52 card deck contains 16 cards with a point value of 10 (jacks, queens, kings and tens) and four aces.a. What is the probability that both cards dealt are aces or 10-point

cards?b. What is the probability that both cards are aces?c. What is the probability that both of the cards have a point value of

10?d. A blackjack is a 10-point card and an ace for a value of 21. Use

your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack.

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Answer to Example 49

a. We have N=52, r=20, n=2 and x=2. So we get, P(x=2)=0.143288b. We have N=52, r=4, n=2 and x=2. So we get, P(x=2)=0.004525c. We have N=52, r=16, n=2 and x=2. So we get, P(x=2)=0.090498

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Solution

d. P(Both cards dealt are either aces or 10 point cards) = P(Both cards are aces OR one card is ace and the other is a 10-point card OR both cards are 10-point cards)= P(Both cards are aces)+P(one card is ace and the other is a 10-point card) +P(both cards are 10-point cards)• Therefore, 0.143288=0.004525+P(Blackjack)

+0.090498• So, P(Blackjack)=0.143288-0.004525-

0.090498=0.048265

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Alternate solution

d. P(Blackjack)= P((first card is an ace AND second card is a 10 point card) OR(first card is a 10-point card AND second card is an ace))==0.076923*0.313725+0.307692*0.078431=0.024133+0.024133

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Continuous Probability Distribution

• The uniform probability density function

= 0 elsewhere Hence c = 1/(b-a)

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The mean & variances

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Example 1

• The random variable x is known to be uniformly distributed between 1.0 and 1.5.

a. Show the graph of the probability density function.

b. Compute P(x=1.25)c. Compute d. Compute

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Solution

a.

1/(1.5-1.0)

0.5 1.0 1.5

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Solution

b. P(x=1.25)= 0c.

d.

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Example 3• Delta Airlines quotes a flight time of 2 hours, 5 minutes

for its flights from Cincinnati to Tampa. Suppose we believe that actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes.

a. Show the graph of the probability density function for flight time.

b. What is the probability that flight will be no more than 5 minutes late?

c. What is the probability that flight will be more than 10 minutes late?

d. What is expected flight time?

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The Normal Distribution

• The probability density function

• Standard Normal Density Function

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The Standard Normal Distribution

• The Normal Curve:

-3 -1.95 -0.90000000000001 0.14999999999999 1.19999999999999 2.249999999999980

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

P(x)

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Example 9

b. To find f(45<x<55) 1. =NORMDIST(55,50,5,TRUE)-NORMDIST(45,50,5,TRUE)Alternatively, 2. Convert to z-score:

=2*0.3413= 0.6826

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Example 11

a. P(-1<=z<=0)=0.3413b. P(-1.5<=z<=0)= P(0<= z<=1.5)=0.4332c. P(-2< z <0)=0.4772d. P(-3< z <0)=0.4986

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Example 15

a. The Area to the left of z is 0.2119 find k. Obviously k is negative. That is to find 0.50-P(k<z<0) That is to find k such that P(k<z<0)=0.50-0.2119=0.2881 So, k=-0.80

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Example 15

b. The area between –z and z is 0.9030Hence Therefore That is Or Alternatively,

Hence,

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Example 15

c. The area between –z and z is 0.2052.Hence the value of k 0.26.d. The area to the left of z is 0.9948.Hence k is in positive region.OR Hence k=2.56e. The area to the right of z is 0.6915.Hence k is in negative region.Hence Therefore -k=-0.50