Chapter 4-Distributions Of Random Variables

Question 1

Normal Curve or Normal Distribution

Answer 1

Describes a symmetric bell curve

Noted as (N(µ,SD))

Many variables are nearly normal but none are exactly normal

Changing the mean shifts the bell curve left or right, accordingly
Change the standard deviation will stretch or compress the bell curve

Question 2

Parameters

Answer 2

The mean and standard deviation of a Normal Distribution

Mean=mu
Standard deviation =sigma

Question 3

Standard Normal Distribution

Answer 3

When the mean =0 and the standard deviation = 1

Question 4

Z-Score

Answer 4

How many standard deviations away from the mean is an observation.

Used to determine how unusual an observation is, when the absolute value of z-score for x1 is greater than the absolute value of z-score for x2, then x1 is said to be more unusual than x2

Mathematically:
Z = (x-µ)/SD

Where,
x=observation
µ=mean/expected value
SD=standard deviation = √Var(x)

Question 5

Solving normal probability examples

Answer 5

1) always draw and label a picture of the Normal Distribution. This will help give estimates
2) calculate the z-score

3) use statistical software to compute the area above or below the z-score accordingly
* Note- many statistical programs give areas to the left only. To compute the right area, subtract value from one (1-left area)

Question 6

68-95-99 Rule

Answer 6

Rule of thumb for the probability of falling within 1, 2, or 3 standard deviations

Question 7

Bernoulli Distribution or Bernoulli Random Variable

Answer 7

When an individual trial has only two possible outcomes

Labeled either 1 (i.e. success) or 0 (i.e. failure)

mu = p
sigma = sqrt(p(1-p))

Where,
p= probability of success
1-p= probability of failure or 0

Question 8

Sample proportion

Answer 8

Is the sample mean of observations

Question 9

Geometric Distribution

Answer 9

Describes the waiting time until a success for independent and identically distributed (iid) Bernoulli random variables

Probability is finding success in nth trial:
((1-p)^(n-1))*p

Where 1-p = probability of failure

I.e. Calculate the probability of rolling a 6 for the first time on the 6th roll of a die?

P(6 on the 6th roll) = (1-(1/6))^(6-1)(1/6) = (5/6)^5(1/6) ≈ 0.067

Question 10

Geometric Distribution: mean/expected value

Answer 10

µ = 1/p

Question 11

Geometric Distribution: Variance

Answer 11

SD^2 = (1-p)/p

Question 12

Geometric Distribution: standard deviation

Answer 12

sigma = √(1-p)/p^2

Question 13

Binomial Distribution

Answer 13

Used to describe the number of successes in a fixed number of trials

The probability of having k successes in n independent Bernoulli trials with probability of success p

(n choose k)p^k(1-p)^(n-k)=(n!)/(k!(n-k)!)p^k(1-p)^(n-k)

To be binomial, it must meet four conditions

1) trials are independent
2) number of trials, n, is fixed
3) each trial outcome can be classified as success or failure
4) probability of a success, p, is the same for each trial

Question 14

Binomial distribution: mean, variance, standard deviation

Answer 14

mu = n*p

sigma^2 = np(1-p)

sigma = sqrt[n*p(1-p)]

Question 15

Completing Binomial Probabilities

Answer 15

1) check that the model is appropriate
2) identify n, p, k
3) use software or the formula to determine the probability and interpret the results

Question 16

Normal approximation of the binomial Distribution

Answer 16

The binomial Distribution with Probability of success p is meant normal when the sample size n is sufficiently large that np and n(1-p) are both at least 10. The approximate Normal Distribution has parameters corresponding to the mean and standard deviation of the binomial Distribution:

mu = n*p
sigma = sqrt[n*p(1-p)]

May need to add extra area (0.5 on either end) when applying the Normal Distribution when examining a range of observations

Question 17

Negative Binomial Distribution

Answer 17

Describes the probability of observing the kth successes on the nth trial

Must meet 4 conditions:

1) trails are independent
2) each trial outcome can be classified as success or failure
3) the probability of success, p, is same for each trial
4) the last trial must be a success

P(the k-th successes on the the n-th trial) = ((n-1)!/(k-1)!)p^k(1-p)^(n-k)

The value p represents the probability that an individual trial is a success

Question 18

Poison Distribution

Answer 18

Useful for estimating the number of events in a larger population over a unit of time, assuming observations/individuals within the population are independent

rate = lambda

P(observe k events) = (lambda^k*e^(-k))/k!

General guidelines:

• population must be large
• events occur independently of each other

Question 19

Computing Percentiles

Answer 19

Using R-studio: pnorm(1800, mean = 1500, sad = 300) where 1800 represents the variable/value in question

Using tables, six-sigma