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GWU / Statistics / STAT / What is the additive rule of probability?

What is the additive rule of probability?

What is the additive rule of probability?

Description

School: George Washington University
Department: Statistics
Course: Introduction to Statistics in Social Science
Term: Spring 2019
Tags: Statistics and midterm
Cost: 50
Name: STAT 1053 - Midterm Study Guide
Description: This study guide contains: - Definitions - Formulas - Examples - Past quizes : questions and answers
Uploaded: 03/14/2019
14 Pages 66 Views 5 Unlocks
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Central Tendency: 


What is the additive rule of probability?



● Measured by mean & median

Mean: 

● A measure of central tendency

● = average

→ + all values together and ÷ by the # of numbers X : sample mean

μ : population mean

n

μ = ∑ i=1

xi 

Median: 

● A measure of central tendency

● Middle value

→ put all numbers in increasing order and ÷ by 2 to determine the middle value X : sample median

μ : population sample median

(n+1)


What is discrete random variable?



Formula for calculating median = 2 

Range: 

● Measure of variability

● biggest value - smallest value

→ 1, 2, 4, 8 ⇒ 8-1=7 ( 7 is the range )

Mode: 

● the most frequently appearing number in a set Variance: 

2

(x −x )

2 ∑ n−1 Don't forget about the age old question of What can a newborn see most clearly before their vision develops?

● s =

i

● σ2: population variance


What is the bbnomial formula?



● σ : population standard deviation

Standard Deviation: 

● s= √s2 

Probability: 

● The extent to which an event is likely to occur measured by a ratio of the favorable cases to the whole number of cases possible

A Sample Point: 

● Is the most basic outcome of an experiment

The Sample Space: 

● The collection of all points of the experiment

An event: 

● A specific collection of sample points

P(A ⋃ B ) → UNION 

● if either A or B ( or both ) occurs

P( A ⋂ B ) → INTERSECTION 

● if both A & B occur on a single performance of the experiment

Conditional Probability: We also discuss several other topics like Do we have case morphology in english?

( A⋂B )

● P(A | B) = provided that P(B) > 0 P (B) 

( A⋂B )

● P(B | A) = provided that P(A) > 0 P (A) 

Complement: 

● the event that A does not occur

c

● A is a complement to an event A

c

● P(A) + P( A ) = 1

○ P(A) = 1 - P( A )c 

Additive Rule of Probability: 

● The union of 2 events A & B:

○ P( A ⋃B ) = P (A) + P(B) - P( A ⋂B ) We also discuss several other topics like What punishment can come with a class c misdemeanor?

■ If not mutually exclusive

Multiplicative Rule of Probability: 

● P( A ⋂B ) = P (A) x P(B | A)

● P( A ⋂B ) = P (B) x P(A | B)

Mutually Exclusive events: 

● If one event happens the other one does not ● Cannot occur at the same time

● P( A ⋂B ) = 0

● P( A ⋃B ) = P(A) + P(B)

Independent events: 

● Events A and B are independent events if the occurrence of B does not alter the Probability that A has occurred If you want to learn more check out What is a species?

● P( A ⋂B ) = P(A) x P(B) If you want to learn more check out Was the pre- wwi era a “golden age”?

● P(A | B) = P(B)

● P(B | A) = P(A)

SUMMARY

Don't forget about the age old question of What characteristics define an animal?

Quiz 2: Questions & Answers:

1. Suppose that each college student is classified as an underclassman or as an upperclassman. A college student is selected at random. The probability that the student is an upperclassman is 0.69. The probability that the student is a male underclassman is 0.14. The probability that the student is male is 0.51.

a. Find the probability that the student is a male or an underclassman ( or both) -

Upperclassman(U)

Underclassman (N)

Male (M)

0.37

0.14

0.51

C)c

Male (M

0.32

0.17

0.49

0.69

0.31

1

- P(M ⋂ N) = P(M) + P(N) − P(M ⋂ N) ==> (0.51 + 0.31) − 0.14 = 0.68

2. The probability that a randomly selected rhino in an animal preserve is grey is 0.74. Suppose that the conditional probability that a rhino is one-horned, given that rhino is grey is 0.40. a. Find the probability that a randomly selected rhino in the animal preserve is grey and one-horned.

- P (G) = 0.74

- P(H|G) = 0.40

- P(H ⋂ G) = P(H|G) x P(G) ⇒ 0.40 x 0.74 = 0.296

● A Random Variable: 

○ a variable that takes numerical values associated with the outcome of an

experiment where one ( and only one) numerical value is assigned to each simple point

■ Discrete

■ Continuous

● Discrete Random Variable: 

○ A random variable that can assume a countable number of value

■ # of households in a group of 1200 who tune in to a particular TV show:

0, 1, 2, …, 1200

● Continuous random variable: 

○ Can assume values corresponding to any of the points contained in an interval ■ The amount of fluid of soda loaded into a 16 fluid ounce bottle

● The Probability Distribution:

○ The specification of the P of each of the possible values of the discrete random variable

● The probability that x takes a particular value has 2 factors: 

○ The probability of a particular outcome

○ The sequence that has the appropriate # of the Heads and Tails

● Binomial Formula: 

r n−r = pr − pn−r 

P(X = r) = p q (1 )

μ = Σ(x) = Σ xp (x)

2 = √Σ(x − μ )2 = Σ − μ2 

σ (x )p(x)

σ = √σ2 

● Binomial Probability Distribution 

○ Mean : Σ(x) = μ = np

2npq = np(1 − p)

○ V ariance (x) : σ =

○ Standard deviation = σ = √npq

● Standard Normal Distribution 

○ Random variable Z

○ μ = 0

○ σ = 1

■ From Normal → Standard

x−μ

● z = σ 

● ! Ti-84 ⇒ 2nd → STAT→ if you need to find the area: ○ between 2 given values ⇒ Normalcdf

○ 1 value ⇒ Normalpdf

Quiz 5: Questions & Answers

1. Suppose that the scores on an academic achievement test are normal, with Mean 650 and Standard Deviation 150. Consider the score of a test-taker selected at random. a. Find the probability that the score is between 575 and 950

- μ = 650

- σ = 150

P ( 575 < X < 950) = P ( - 0.5 < Z < 2) = 0.6687

575−650 = − 0

→ Z = .5 150 

950−650 =

→ Z = 2 150 

b. What is the t-score such that the probability is 0.13 that a randomly selected person get a score less than t?

- μ = 650

- σ = 150

P ( X < t )= 0.13

t−650 

P ( Z < ) = 0.13 < 0.5 t 

t−650 < 0

150

−(t−650) = 0

P ( Z > ) .13 150 

(t−650) = 0 − 0 = 0

P ( 0 < Z < .5 .13 .37 150 

−t−650 = 1

.13 150 

t = 480.5

Quiz 6: Questions & Answers

1. Suppose that the Population Mean score on a test is 600 points and the Standard Deviation is 130 points. A random sample of 250 test takers is selected.

a. Find the probability that the Sample Mean score is between 590 and 605 points. - X =600

- σ = 130

- σ = =8.222 xσ√n

590−600 ≤ Z ≤ 8.222 

605−600 = P 1 ≤ Z ≤ 0 = 0

- P( 590 ≤ x ≤ 605) = P( ) (− .22 .608) .617 8.222 

2. Suppose that 30% of all US college student major in the arts or sciences. A Random Sample of 500 US college students is selected.

a. Find the probability that fewer than 165 of them major in arts and sciences. - p=0.3

- n=500

- Find P(x<165)

- μ = np = 150

- σ = √np(1 − p) = √150 • 0.7 = 10.25 

165−150 = P < 1 = 0

- P(X < 165) = P(Z < ) (Z .46) .9279 10.25 

● In a sample of n, the Sample Proportion of success is: 

○ p︿= nx 

■ Where x is the number of successes of a binomial variable

● Parameter:

○ A numerical descriptive measure of a population (ex: σ, μ, p)

● Sampling Distribution for x : 

○ The value varies from sample to sample

○ It is a random Q

μx = μ

σx =σ√n

n: sample of size n from a population

● For sample statistics, as a random variable:

○ Expected value

○ Variance

○ Standard deviation

● Confidence Interval (CI): 

○ The confidence in the estimation process rather than in the particular interval that is calculated from sample data

○ It explains that over many repetitions of this application using the same procedure, …% of the resulting intervals will contain μ

● Example: CI 90% → 0.9

○ (1- α) = 0.9

○ α = 0.1

○ .05 → look at the z-table (above) or 2α = 0

■ Ti-84: invNorm⇒ a = 0.05 , μ = 0, σ = 1

○ Z = 1.645

α

2

■ For both sides of the standard distribution curve → -1.645, 1.645

● Example:

○ n=1000

︿= =

○ p 73% 0.73

■ p ± Z

α√ np·q 2

■ σp = √ np·q

● q = 1 - p = 1 - 0.73 = 0.27

○ From the normal area table values at 0.05level of significance ( for CI 95%) at 0.05 level

of significance, Z = 1.96

α

2

■ Plugging in the numbers and getting CI 95% (0.7025, 0.7575)

● Ti-84: Stat → 1propZInt + !for calculator you need x value, which is: ○ p · n = x ⇒ x=730

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