27
Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random variable. 6.1b h.w: pg pg 354: 14, 18, 19, 23, 25

Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Embed Size (px)

Citation preview

Page 1: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Discrete and Continuous Random VariablesI can find the standard deviation of discrete random variables. I can find the probability of a continuous random variable.

6.1b

h.w: pg pg 354: 14, 18, 19, 23, 25

Page 2: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

The Variance of a Discrete Random Variable

Recall:

Variance and standard deviation are measures of spread.

Page 3: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

If X is a discrete random variable with mean μ, then the variance of X is

The standard deviation is the square root of the variance.

2

1

2 222 21

2

X X k X k

i i

X

X

x p x p x p

x p

Page 4: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Example: Selling of Aircraft Gain Communication sells aircraft

communications units to both the military and the civilian markets. Next years sales depend on market conditions that can not be predicted exactly.

Page 5: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Gains follows the modern practice of using probability estimates of sales.

The military estimates the sales as follows: Units sold:1000 3000 5000 10,000 Probability: 0.1 0.3 0.4 0.2

Take X to be the number of military units sold.

Page 6: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Compute μx:

μx

= (1000)(0.1) + (3000)(0.3) + (5000)(0.4) + (10000)(.2)

= 100 + 900 + 2000 + 2000

= 5000 units

Page 7: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Calculate the variance of X:

σx2= ∑(xi - μx)2 pi

= (1000 – 5000)2(0.10) + … finish

= 7,800,000

Page 8: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Standard Deviation σx = sqrt 7,800,000

= 2792.8

The standard deviation is the measure of how variable the number of units sold is.

Page 9: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

To find the variance with calculator:

In notes for your info. Try it if you want on your own.

Page 10: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Recall: If we use a table or a calculator to select digits 0 and 1, the result is a discrete random variable which we can “count”.

What is the probability of 0.3 ≤ X ≤ 0.7 ? Infinite possible values!

Page 11: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Now we will assign values as areas under a density curve.

Page 12: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Continuous Random Variables

A continuous random variable X takes all values in a given interval of numbers.

The probability distribution of a continuous random variable is shown by a density curve.

Page 13: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

The probability that X is between an interval of numbers is the area under the density curve between the interval endpoints.

The probability that a continuous random variable X is exactly equal to a number is zero

Page 14: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Example: Uniform Distribution(of random digits between 0 and 1)

Note: P(X ≤ 0.5 or X ≥ 0.8)

= 0.7 We can add non-overlapping parts.

Page 15: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Normal Distributions as Probability Distributions

Recall N(μ,σ) is the shorthand notation for the normal distribution having mean μ and standard deviation σ.

Page 16: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

If X has the N(μ,σ) then the standardized variable

Z = (X – μ) / σ

is a standard normal random variable having the distribution N(0,1).

Page 17: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Example: Drugs In School An opinion poll asks a SRS of 1500 of U.S.

adults what they think is the most serious problem facing our schools.

Suppose 30% would say “drugs.” The population parameter p

is approximately N(0.3, .0118).

Page 18: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

What is the probability that the poll differs from the truth about the population by more than 2 percentage

points? More than one way to do this.

= P(p < 0.28 or p > 0.32) The”shaded region”

= P(p < 0.28) + (p > 0.32)

Page 19: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Confirm the z-scores and the area of the shaded region.

Standardize the values.

P(p < 0.28) = P( z < (0.28 – 0.30)/0.0118 ) = P(z < -1.69)

Page 20: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Use z-score to find the area.

Calc:

2nd VARS(DIST):normalcdf(-EE99, -1.69)

= 0.0455

Page 21: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

P(p > 0.32) = 0.0455 also why? P(p < 0.28) + (p > 0.32) = 0.0455 + 0.0455

= 0.0910 Conclusion:

The probability that the sample will miss the truth by more than 2 percentage points is 0.0910.

Page 22: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Or, use the complement to get “middle” or “unshaded” region:

1 – P(-1.69 < z < 1.69) = 1 - normcdf(-1.69,1.69)

= 1 - 0.9090 = .0901

with complement rule

Page 23: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Exercise: Car Ownership Chose an American at random and let the

random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars.

Page 24: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Probability model

a) Verify that this is a legitimate discrete distribution.

Number of cars X

0 1 2 3 4 5

Prob. 0.09 0.36 0.35 0.13 0.05 0.02

Page 25: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

Display the distribution in a probability histogram. (2 min)

Page 26: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

b) Say in words what the event {X ≥ 1} is.

The event that the household owns at least one car.

Find P(X ≥ 1) = P(X = 1) + P(X = 2) + … + P(X=5)

= 0.91

Or, 1 – P(X = 0) = 1 – 0.09

= 0.91

Page 27: Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random

c) A housing company builds houses with two-car garages.

What percent of households have more cars than the garage can hold?

P(X > 2) = P(X=3) + P(X=4) + P(X+5)

= 0.20

20% of households have more cars than the garage can hold.