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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

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**The Variance of a Discrete Random Variable**

Recall: Variance and standard deviation are measures of spread.

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**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.

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**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.

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

The military estimates the sales as follows: Units sold: ,000 Probability: Take X to be the number of military units sold.

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**Compute μx: μx = (1000)(0.1) + (3000)(0.3) + (5000)(0.4) + (10000)(.2)**

= = 5000 units

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**Calculate the variance of X:**

σx2= ∑(xi - μx)2 pi = (1000 – 5000)2(0.10) + … finish = 7,800,000

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

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**To find the variance with calculator:**

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

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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!

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

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**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.

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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

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**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.

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**Normal Distributions as Probability Distributions**

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

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**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).

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**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).

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**= 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)

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

Standardize the values. P(p < 0.28) = P( z < (0.28 – 0.30)/ ) = P(z < -1.69)

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**Use z-score to find the area.**

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

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P(p > 0.32) = also why? P(p < 0.28) + (p > 0.32) = = Conclusion: The probability that the sample will miss the truth by more than 2 percentage points is

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**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) = = .0901

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**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.

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**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.

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

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**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

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**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.

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