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Probability Models for Distributions of Discrete Variables

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Determine the probability a student’s number of siblings falls within 1 standard deviation of the mean.

Determine the probability a student’s number of siblings falls within 2 standard deviations of the mean.

Determine the probability a student’s number of siblings falls within 2 standard deviations of the mean.

Randomly select a college student. Determine x, the number of credit cards the student has.

x = # of cards p(x) = probability of x occurring

A population is a collection of all units of interest.

Example: All college students

A sample is a collection of units drawn from the population.

Example: Any subcollection of college students.

Probabilities go with populations.

Scientific studies randomly sample from the entire population.

Each unit in the sample is chosen randomly.

The entire sample is random as well.

Populations / Samples

For discrete data, a population and a sample are summarized the same way (for instance, as a table of values and accompanying relative frequencies).

A probability distribution (or model) for a discrete variable is a description of values, with each value accompanied by a probability.

Probability Models and Populations

2. the probability of an event is the long term (technically forever) relative frequency of occurrence of the event, when the experiment is performed repeatedly under identical starting conditions.

3. The probability of an event is the relative frequency of units in the population for which the event applies.

To aggregate these meanings:

The probability associated with an event is its relative frequency of occurrence over all possible ways the phenomena can take place.

Probability Models and Populations

A probability distribution for a discrete variable is tabulated with a set of values, x and probabilities, p(x).

Probabilities

Must be nonnegative.

A probability distribution for a discrete variable is tabulated with a set of values, x and probabilities, p(x).

- Probabilities
- Must be nonnegative.
- Must sum to 1.
- Within rounding error.

The mean of a probability distribution is the mean value observed for all possible outcomes of the phenomena.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

Mean = 1.80 SD = 1.44

0 0 0 0 0 0 0 … 0 (200)

1 1 1 1 1 1 1 1 1 1 … 1 (300)

2 2 2 2 2 2 … 2 (200)

3 3 3 3 … 3 (150)

4 4 … 4 (100)

5 … 5 (50)

Mean = 1.80 SD = 1.44

Values for the mean and standard deviation don’t depend on the number of data values; they depend instead on the relative location of the data values – they depend on the distribution in relative frequency terms.

The mean of a probability distribution is the mean value observed for all possible outcomes of the phenomena.

Formula:

is synonymous with “population mean”

SUM symbol

Greek letter “myou”

The standard deviation of a probability distribution is the standard deviation of the values observed for all possible outcomes of the phenomena.

Formula:

denotes “population standard deviation”

Greek letter “sigma”

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

Mean = 1.80 SD = 1.44

Mean – SD = 0.56 Mean + SD = 3.24

65 / 100 = 65%

x = # children in randomly selected college student’s family.

0.2194 = 21.94% of all college students come from a 1 child family.

Guess at mean? Above 2

(right skew mean > mode).

To determine the mean, multiply values by probabilities,

xp(x)

and sum these.

55/10 = 5.50 is not the mean

1.000/10 = 0.10 is not the mean

To determine the variance, multiply squared deviations from the mean by probabilities,

(x – )2p(x)

and sum these.

The standard deviation is the square root of the variance.

Examining the data set consisting of # of children in the family recorded for all students: The mean is 2.743; the standard deviation is 1.468.

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5)

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5)

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5)

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5)

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5)

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5) = 0.0317

+ 0.0124

+ 0.0043

+ 0.0005

+ 0.0003

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5) = 0.0317

+ 0.0124

+ 0.0043

+ 0.0005

+ 0.0003

= 0.0492

Determine the probability a student is from a family with more than 5 siblings.

P(x > 5) = 0.0492

4.92% of all college students come from families with more than 5 children (they have 4 or more brothers and sisters).

Determine the probability a student is from a family with at most 3 siblings.

P(x 3) = 0.2194

+ 0.2806

+ 0.2329

= 0.7329

Determine the probability a student is from a family with at least 7 siblings.

P(x 7) = 0.0124

+ 0.0043

+ 0.0005

+ 0.0003

= 0.0175

Good idea: Take the reciprocal of a small probability…

1/.0175 = 57.1 1 in 57 students

Determine the probability a student is from a family with fewer than 5 siblings.

P(x< 5) = 0.2194

+ 0.2806

+ 0.2329

+ 0.1442

= 0.8771

less than or equal to 3 greater than or equal to 7

no more than 3 no fewer/less than 7

x 3x 7

Determine the probability a student’s number of siblings falls within 1 standard deviation of the mean.

Guess?

0.68

Determine the probability a student’s number of siblings falls within 1 standard deviation of the mean.

Mean= 2.743

SD = 1.468

1 SD below the mean

2.743 – 1.468 = 1.275

1 SD above the mean

2.743 + 1.468 = 4.211

Determine the probability a student’s number of siblings falls within 1 standard deviation of the mean.

1 SD below the mean = 1.275

1 SD above the mean = 4.211

Values are within 1 SD of the mean if they are between these.

Determine the probability a student’s number of siblings falls within 1 standard deviation of the mean.

1 SD below the mean = 1.275

1 SD above the mean = 4.211

Values are within 1 SD of the mean if they are between these.

1 SD below the mean = 1.275

1 SD above the mean = 4.211

Values are within 1 SD of the mean if they are between these.

The probability of being between these:

0.2806 + 0.2329 + 0.1442 = 0.6577

Determine the probability a student’s number of siblings falls within 2 standard deviations of the mean.

Guess? 0.95

2 SD below the mean

1.275 – 1.468 = -0.193

2 SD above the mean

4.211+ 1.468 = 5.679

Between -0.193 and 5.679.

Determine the probability a student’s number of siblings falls within 2 standard deviations of the mean.

Between -0.193 and 5.679.

(Equivalent to 5 or fewer.)

Determine the probability a student’s number of siblings falls within 2 standard deviations of the mean.

Between -0.193 and 5.679.

(Equivalent to 5 or fewer.)

We know an outcome more than 5 has probability 0.0492.

Between -0.193 and 5.679.

(Equivalent to 5 or fewer.)

We know an outcome more than 5 has probability 0.0492.

The probability of an outcome at most 5 is 1 – 0.0492 = 0.9508.

Between -0.193 and 5.679.

0.9508.

A company monitors pollutants downstream of discharge into a stream.

Data were collected on 200 days from a point 1 mile downstream of the plant on Stream A.

Data were collected on 100 days from a point 1 miles downstream of the plant on Stream B.

Pollutant Particles in Streamwater

Here is the probability distribution for the number of diners seated at a table in a small café.

a) Fill in the blank

Here is the probability distribution for the number of diners seated at a table in a small café.

a) Fill in the blank

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Start by computing xp(x) for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Start by computing xp(x) for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Start by computing xp(x) for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Start by computing xp(x) for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Start by computing xp(x) for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Sum these.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the mean
- Sum these.
- = 3.00

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- ( x – ) 2p(x)
- for each row.

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- ( x – )2p(x)
- for each row.
- = 3

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- ( x –3)2p(x)
- for each row.
- = 3

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- ( x –3) 2p(x)
- for each row.
- = 3

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- ( x –3) 2p(x)
- for each row.
- = 3

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Start by computing
- (x–3) 2p(x)
- for each row.
- = 3

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Sum these

Here is the probability distribution for the number of diners seated at a table in a small café.

- b) Determine the standard deviation
- Sum these
- Variance = 1.00
- SD: = 1.00

This framework makes it possible to obtain fairly good approximations to means and standard deviations from a histogram of continuous data.

[Optional] Application

Here are waiting times between student arrivals in a class. There are 21 students (20 waits).

Example

Approximate the mean and median. How do they compare?

For each class, determine its frequency and corresponding midpoint.

Example: Mean

Frequency = 10

Midpoint = 5

Tabulate frequencies and midpoints.

Example: Mean

Tabulate frequencies and midpoints.

Example: Mean

Obtain relative frequencies.

Example: Mean

Obtain relative frequencies.

Example: Mean

Proceed with the formula

Example: Mean

Find the value with 50% below and 50% above.

Example: Median

Obtain relative frequencies.

Example: Median

Find the value with 50% below and 50% above.

Example: Median

10 of 20 = 50% below 10

Median 10.00

Mean 14.00

Range 44

S.D. 11

1.3 1.9 1.9 2.5 2.6 3.0 3.6 3.7 5.9 9.7 10.4 10.6 11.2 13.5 15.9 21.4 27.5 29.8 33.6 43.5

Approximations: Actual Values:

Median 10.0.05 Median =

Mean 14.0 Mean =

Range 44 Range =

SD 11 SD =

Example: Data / Exact Values

1.3 1.9 1.9 2.5 2.6 3.0 3.6 3.7 5.9 9.7 10.4 10.6 11.2 13.5 15.9 21.4 27.5 29.8 33.6 43.5

Approximations: Actual Values:

Median 10.0.05 Median = 10.05

Mean 14.0 Mean = 12.68

Range 44 Range = 42.2

SD 11 SD = 12.31

Example: Data / Exact Values

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