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)