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

The Variance of a Discrete Random Variable Recall: Variance and standard deviation are measures of spread.

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.

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.

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.

Compute μx: μx = (1000)(0.1) + (3000)(0.3) + (5000)(0.4) + (10000)(.2) = 100 + 900 + 2000 + 2000 = 5000 units

Calculate the variance of X: σx2= ∑(xi - μx)2 pi = (1000 – 5000)2(0.10) + … finish = 7,800,000

Standard Deviation σx = sqrt 7,800,000 = 2792.8 The standard deviation is the measure of how variable the number of units sold is.

To find the variance with calculator: In notes for your info. Try it if you want on your own.

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!

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

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.

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

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.

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

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).

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).

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

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)

Use z-score to find the area. Calc: 2nd VARS(DIST):normalcdf(-EE99, -1.69) = 0.0455

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.

Or, use the complement to get “middle” or “unshaded” region: with complement rule 1 – P(-1.69 < z < 1.69) = 1 - normcdf(-1.69,1.69) = 1 - 0.9090 = .0901

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.

Probability model Number of cars X 1 2 3 4 5 Prob. 0.09 0.36 0.35 0.13 1 2 3 4 5 Prob. 0.09 0.36 0.35 0.13 0.05 0.02 a) Verify that this is a legitimate discrete distribution.

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

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

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.