MAT 209 - Statistics Chapter 6 - Probability Distributions

THE FOLLOWING NOTES USE EXTENSIVE MS Word Math Equations that will NOT transfer over to HTML.

Chapter 6 Probability Distributions

Random variable = Quantities that can take on different values depending on chance.

·      Discrete random variable – Finite # of values (Ex: Dice)

·      Countable infinity of values – As many values as whole numbers

Probability Distribution assigns a probability to each measurable subset of the possible outcomes of a random experiment, survey, or procedure of statistical inference.

6.1 Probability Distributions

Probability distribution of random variable (single die) example assigns probability to values:

Number rolled on Die	Probability
1	1/6
2	1/6
3	1/6
4	1/6
5	1/6
6	1/6

Probability distribution written as function:

Two general rules for probability distribution:

1.     Values of probability distribution must be numbers between 0 and 1 inclusive.

2.     A random variable has to take on one of its values, sum of all values = 1.

6.2 Binomial Distribution

Borrowing the language of games of chance for the probability that an event will occur x times out of n

We assume that for this section …

1.     There are a fixed number of trials resulting in success/failure.

2.     The probability of success is the same for each trial.

3.     Trials are all independent.

Example p.203 Student reduced to guessing on 8 questions on a multiple-choice test. Each question has 5 choices, but ONLY one is correct. If C = correct answer and I = incorrect answer what are probability that the student will get 6 questions right and 2 questions wrong.

One possible outcome is CCCCCCII.

Since P(C) = 1/5 and P(I) = 4/5 à CCCCCCII = (⅕)(⅕)(⅕)(⅕)(⅕)(⅕)(⅘)(⅘) = (⅕)⁶(⅘)²

However, the CCCCCCII is only one possibility à Need to count the ways of arranging 6 C’s and 2 I’s à Combination  à multiply by probability of 6 C’s and 2 I’s à

Binomial Distribution: If n = # of trials, p = probability of success of each trial, and all trials are independent à probability of x successes in n trials is …

Jargon: Number of successes in n trials is a random variable having binomial probability distribution/binomial distribution.

Example 6.5 Calculate the probability that 2 of 5 reviewers will like a new movie, given that the probability that any one of them will like is 0.70. Plug in for binomial distribution formula.

X = 2, n = 5, p =0.70

Sampling a binomial population = When we observe a value of a random variable having the binomial distribution.

6.3 Hypergeometric Distribution

Different from binomial distribution – selection with NO replacement.

If n objects are chosen at random from a set consisting of a objects of type A and b objects of type B, and the selection is without replacement à probability of “x of type A and n – x of type B” is …

Example 6.10 A department store has 16 delivery trucks, 5 have worn brakes. If 3 randomly picked for general overhaul, find the separate probabilities that these 3 will include 0,1,2, or 3 trucks with worn brakes.

Problem has sampling without replacement. Therefore 5 trucks (a) have worn brakes (Type A = Trucks with worn brakes). 11 trucks (b) are fine without worn brakes (Type B = Trucks w/o worn brakes). 3 Random trucks taken (n). Chance of success or chance of getting 0, 1, 2, or 3 trucks (x) with worn brakes in the random sample of 3.

f(0) = (1*165)/560 ≈ 0.295

f(1) = (5*55)/560 ≈ 0.491

f(2) = (10*11)/560 ≈ 0.196

f(3) = (10*1)/560 ≈ 0.018

*Binomial distribution & Hypergeometric distribution values are close enough to each other (as demonstrated in the textbook example 6.13) when certain conditions are fulfilled:

Sample does NOT exceed 5% of population à .

Approximation of binomial distribution and hypergeometric distribution is useful b/c binomial distribution is less complicated (binomial has two parameters n & p vs. hypergeometric’s three a, b, and n.

6.4 Poisson Distribution

When n (number of trials) is large and p (probability of trial success) is small (when n≥100 and (n)*(p)<10), binomial distribution changes to
Poisson distribution approximation to binomial distribution:

Poisson Distribution useful b/c it has only one parameter (np) vs binomial distribution (2) and hypergeometric distribution (3).

Also has a generalized Poisson Distribution (with parameter λ):

 Where, λ = expected/average number of successes, can be applied to situations where we expect a fixed number of successes per unit time (or some other kind of unit).

Example 6.14 If 1% of labels in a large shipment of jars are defective, use the Poisson distribution to find the probability that among 400 randomly inspected jars, only 3 will have defective labels.

X = 3   n = 400          p = 0.01          (n)(p) = 4      e^-4 = 0.018

f(3) = (0.018)(4³/3!) = 0.1920

There are cases where the hypergeometric distribution ≈ Poisson distribution.
NO replace + large n values.

Example 6.16 Auditor asked to investigate a collection of 4,000 sales invoices, of which 28 contain errors. Sample of 150 invoices is selected. What is the probability that this set of 150 invoices will contain exactly 2 with errors?

Hypergeometric problem b/c NO replacement à a = 28, b = 3972, n = 150, and x = 2.

 since, 0.05(4000) = 200 and 150 < 200 à hypergeometric distribution ≈ binomial distribution.

n > 100 and np <10 since, 150 > 100 and (150)(0.007) = 1.05 < 10 à Poisson approximation ≈ binomial distribution.

Example 6.17 Bank receives on average λ = 6 bad checks/day, what is the probability that it will receive 4 bad checks on any given day?

SUMMARY:

Binomial distribution – Repeated independent trials.

Hypergeometric distribution – Sample w/o replacement from a finite population.

Poisson distribution – Counted events w/ NO fixed upper limit.

6.5 Multinomial Distribution

When:

·      More than 2 possible outcomes for each trial

·      Probability of various outcomes remain the same each trial

·      Trials are all independent

Mutlinomial Distribution: If k possible outcomes for each trial and their probabilities are p₁… and p_k, it can be shown that the probability of x₁ outcomes of first kind … and x_k outcomes of the kth kind in n trials. Requirements are that x₁ + x₂ + … + x_k = n   and     p₁ + p₂ + … +p_k = 1.

Example 6.18 City has 3 channels for viewing on Saturday night. Each channel has varying percentage of viewing audiences. Channel 3 has 50% of viewing audience, channel 12 has 30%, channel 10 has 20%. What is the probability that among 8 TV viewers randomly selected in that city on Saturday night, 5 will be watching channel 3, 2 will be watching channel 12, and 1 will be watching channel 10?

N = 8, x₁ = 5, x₂ = 2, x₃ = 1, p₁ =0.50, p₂ = 0.30, and p₃ = 0.20

6.6 Mean Probability Distribution

Mean of Probability Distribution: Multiply the outcome by its probability and add all values.

Where µ = mean of probability distribution, x = random variable, f(x) = probability for x.

*Similar to mean of population related mean (denoted by µ).

Above formula can be further simplified for mean of binomial distributions:

Where µ = mean of probability distribution, n = trials, p = probability of success for each trial.

6.7 Standard Deviation of a Probability Distribution

Similar procedure for measuring variability, BUT instead of averaging square deviations from the mean calculate expected value.

Variance of Probability Distribution:

Where σ = standard deviation, σ² = variance (of probability distribution in this case),

µ = probability distribution mean, f(x) = probability for x, x = random variable.

Standard Deviation of Probability Distribution

Note the similarities to standard deviation of population.

Example 6.6 – Expansion Example 6.22

Probability is 0.60 that a person shopping at a certain market will spend at least $50.

Find probability that among 5 people shopping at this market, 0, 1, 2, 3, 4, or 5 will at least spend $50.

f(0) =  = 0.010

f(1) =  = 0.077

f(2) =  = 0.230

f(3) =  = 0.346

f(4) =  = 0.259

f(5) =  = 0.078

Determine the standard deviation of distribution among the five people who will spend at least $50 at the market.

µ = 0(0.01024) + 1(0.077) + 2(0.230) + 3(0.346) + 4(0.259) + 5(0.078) = 3.001

Number of Persons, x	X - µ	(x-µ)2	Probability f(x)	(x-µ)2*f(x)
0	-3	9	0.010	0.090
1	-2	4	0.077	0.308
2	-1	1	0.230	0.230
3	0	0	0.346	0
4	1	1	0.259	0.259
5	2	4	0.078	0.312

Total σ² = 1.199        σ = (1.199) ^(1/2) = 1.0949885 ~ 1.095

Standard deviation of probability distribution can be simplified when dealing with other types of probability distributions. For example: Standard Deviation of Binomial distribution

where σ = standard deviation of binomial distribution, n = # of trials, p = probability of success.

Example 6.24 Find variance of probability distribution of number of heads obtained in 3 flips of a balanced coin.

n = 3, p = 0.5             1 – p = 0.5

σ² = (3)(0.5)(1-0.5) = 0.75

For hypergeometric distributions the variance is given by .

For Poisson distribution standard deviation is σ =(λ)^½ , with variance being σ² = λ.

6.8 Chebyshev’s Theorem

Variance & standard deviation of probability distribution measure expected size of fluctuation of random variable.

When σ is small à Likely to be close to mean

When σ is large à Likely to far from mean

Chebyshev’s Theorem for probability: The probability that a random variable will take on a value within k standard deviations of the mean is at least 1 – (1/k²).

Therefore the probability of getting a value within 2 standard deviations (2σ) is at least ¾ and the probability of getting a value within 5 deviations (5σ) is at least 24/25.

Example 6.26 What does Chebyshev’s theorem with k =6 tell us about the number of heads that we may get in 400 flips of a balanced coin?

N = 400          p = ½ = 0.5    µ = n*p = 400*0.5 = 200

σ = (n*p*(1-p)) ^½ = (400*0.5*(1-0.5))^½ = 10       à        µ ± 6σ = 140 or 260

Probability of at least 1 – (1/6²) = 35/36 that in 400 flips there will be between 140 and 260 heads.