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Understanding Random Variables: Discrete vs. Continuous

Learn to calculate probabilities for both discrete and continuous random variables, create probability histograms, and understand probability distributions. Explore examples and concepts in AP Statistics.

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Understanding Random Variables: Discrete vs. Continuous

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  1. AP Statistics 7.1 Discrete and Continuous Random Variable

  2. Learning Objective: • Calculate the probability of a discrete random variable and display in a graph. • Calculate the probability of a continuous random variable using a density curve.

  3. Review • We toss a coin 4 times. Our outcome is HTTH. • Let x=# of heads, therefore x=2. • If we got TTTH, then x=1. • The values of x are 0,1,2,3,4. (x is called a random variable)

  4. Definitions • RANDOM VARIABLE: a variable whose value is a numerical outcome of a random phenomenon • DISCRETE RANDOM VARIABLE X: has a countable # of outcomes (possible values)

  5. PROBABILITY DISTRIBUTION: Lists their values and their probabilities. • Two Requirements that the probabilities must satisfy • 1- p has to be between 0 and 1. • 2- p₁ + p₂ + … = 1

  6. Ex. 1: The instructor of a large class gives 15% each of A’s and D’s, 30% each of B’s and C’s, and 10% F’s. Choose a student at random from this class. To “choose at random” means to give every student the same chance to be chosen. The student’s grade on a four point scale (A=4) is a random variable x: • P(Grade is a B or higher) = P(3 or 4)= P(3) +P(4) = 0.3 + 0.15=0.45

  7. PROBABILITY HISTOGRAMS- • used to picture the probability distributions of a discrete random variable. • Create a histogram for the above data:

  8. Ex 2: What is the probability distribution of the discrete random variable X that counts the number of heads in four tosses of a coin? We Know: 1- outcomes ( H or T) 2- independent • There are 16 possible outcomes. • For ex: P(HTTH)= 1/16 • What is P(x=0)= 1/16 (TTTT) • P(x=1)= 1/4 P(x=2)=3/8 P(x=3)= 1/4 P(x=4) = 1/16

  9. Probability Distribution in Table form: • What is the probability of tossing at least 1 head? P(x≥1)=1-P(0)= 1- (1/16)= 15/16 • What is the probability of tossing no more than 3 heads? P(x≤3)= P(x<4)= 1- P(x=4)= 1-(1/16)=15/16

  10. Pg. 373: 7.3 • A) 1% • B) All probabilities are between o and 1. They also add up to 1. • C) P(x≤3)= 0.94 • D) P(x<3)=0.86 • E) P(x≥4)=P(x>3)= 0.06 • F)Let 01-48=class 1,49-86=class2 87-94=class 3, 95-99=class 4, 00= class 5. Use a RDT to repeatedly generate 2 digit #’s to find the proportion of those from 01-94.

  11. Continuous Random Variables • Look at Figure 7.4 pg. 375 (spinner)  • A spinner generates a random number between 0 and 1. • What is the sample space?  S={ 0 ≤ x < 1 } • *We cannot assign probabilities to each individual value of x and then sum, because there are infinitely many possible values. • -Instead we use intervals (area under a density curve)!!! (A new way of assigning probabilities directly to events).  

  12. Ex 3: (Using the spinner) • What is P(.3 < x < .7) = .4 • P(x<.5)= 0.5 • P(x > .8)= 0.2 • P(x < .5 or x > .8)= 0.7 • P(x = .8)= 0 • We call X a continuous random variable because its values are not isolated #’s but an entire interval of #’s

  13. CONTINUOUS RANDOM VARIABLE X- takes all values in an interval of numbers • PROBABILITY DISTRIBUTION- density curve Difference between the two random variables: discrete (specific values) continuous (intervals)

  14. Normal Distributions as probability distributions • N ( μ , σ ) = N(mean, standard deviation) • The standardized variable is Z= (x-µ)/σ

  15. Suppose p is a random variable that has approximately the N(0.3, 0.0118) distribution. • What is P(p< .28 or p> .32) • Step 1: Draw and shade your normal curve:

  16. Step 2: Convert to z-scores • P( p< .28)= P(z< (0.28-0.3)/0.0118)= P(z<-1.69)= 0.0455 • P(p > .32) =P(z> (0.32-0.3)/0.0118)= P(z>1.69)= 0.0455 0.0455+0.0455= 0.091

  17. Complete 7.4-7.5 (pg. 379)

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