Discrete and Continuous Random Variables I can find the standard deviation of discrete random...
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![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.](https://reader036.fdocument.pub/reader036/viewer/2022062318/551ab261550346e0158b638c/html5/thumbnails/1.jpg)
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
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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.
2
1
2 222 21
2
X X k X k
i i
X
X
x p x p x p
x p
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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:1000 3000 5000 10,000 Probability: 0.1 0.3 0.4 0.2
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)
= 100 + 900 + 2000 + 2000
= 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
= 2792.8
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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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)/0.0118 ) = P(z < -1.69)
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Use z-score to find the area.
Calc:
2nd VARS(DIST):normalcdf(-EE99, -1.69)
= 0.0455
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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.
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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
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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
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
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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
= 0.91
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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.