Post on 27-Dec-2015
AP Statistics7.1 Discrete and Continuous Random Variable
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.
Learning Objective:
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)
Review
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)
Definitions
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
PROBABILITY DISTRIBUTION:
Value of X x₁ x₂ x₃ ....
Probability p₁ p₂ p₃ ....
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
Grade 0 1 2 3 4
Prob. 0.10 0.15 0.30 0.30 0.15
used to picture the probability distributions of a discrete random variable.
Create a histogram for the above data:
PROBABILITY HISTOGRAMS-
0 1 2 3 40
0.050.1
0.150.2
0.250.3
0.35
Grades
Grades
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)=3/8P(x=3)= 1/4 P(x=4) = 1/16
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?
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
Number of heads
0 1 2 3 4
Probability 1/16 1/4 3/8 1/4 1/16
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.
Pg. 373: 7.3
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).
Continuous Random Variables
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
Ex 3: (Using the spinner)
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)
N ( μ , σ ) = N(mean, standard deviation)
The standardized variable is Z= (x-µ)/σ
Normal Distributions as probability distributions
What is P(p< .28 or p> .32)
Step 1: Draw and shade your normal curve:
Suppose p is a random variable that has approximately the N(0.3, 0.0118) distribution.
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
Complete 7.4-7.5 (pg. 379)