Chapter 4 part2- Random Variables

INTRODUCTION TO
STATISTICS & PROBABILITY
Chapter 4:
Probability: The Study of Randomness
(Part 2)
Dr. Nahid Sultana
1

Chapter 4
Probability: The Study of Randomness
4.1 Randomness
4.2 Probability Models
4.3 Random Variables
4.4 Means and Variances of Random Variables
4.5 General Probability Rules*
2

4.3 Random Variables
3
 Random Variable
 Discrete Random Variables
 Continuous Random Variables
 Normal Distributions as Probability Distributions

4
Random Variables
4
 A probability model: sample space S and probability for each outcome.
 A numerical variable that describes the outcomes of a chance process is
called a random variable.
 The probability model for a random variable is its probability distribution.
The probability distribution of a random variable gives its possible
values and their probabilities.
Example: Consider tossing a fair coin 3 times.
Define X = the number of heads obtained.
X = 0: TTT
X = 1: HTT THT TTH
X = 2: HHT HTH THH
X = 3: HHH
Value 0 1 2 3
Probability 1/8 3/8 3/8 1/8

5
Discrete Random Variable
Two main types of random variables: discrete and continuous.
A discrete random variable X takes a fixed set of possible values
with gaps between.
The probability distribution of a discrete random variable X lists the
values xi and their probabilities pi:
The probabilities pi must satisfy two requirements:
1. Every probability pi is a number between 0 and 1.
2. The sum of the probabilities is 1.

6
Discrete Random Variable (Cont…)
Example: Consider tossing a fair coin 3 times.
Define X = the number of heads obtained.
X = 0: TTT
X = 1: HTT THT TTH
X = 2: HHT HTH THH
X = 3: HHHValue 0 1 2 3
Probability 1/8 3/8 3/8 1/8
Q1: What is the probability of tossing at least two heads?
Ans: P(X ≥ 2 ) = P(X=2) + P(X=3) = 3/8 + 1/8 = 1/2
Q2: What is the probability of tossing fewer than three heads?
Ans: P(X < 3 ) = P(X=0) +P(X=1) + P(X=2) = 1/8 + 3/8 + 3/8
= 7/8
Or P(X < 3 ) = 1 – P(X = 3) = 1 – 1/8 = 7/8

7
Discrete Random Variable (Cont…)
Example: North Carolina State University posts the grade distributions for its
courses online. Students in one section of English210 in the spring 2006
semester received 31% A’s, 40% B’s, 20% C’s, 4% D’s, and 5% F’s.
The student’s grade on a four-point scale (with A = 4) is a random
variable X. The value of X changes when we repeatedly choose students at
random , but it is always one of 0, 1, 2, 3, or 4. Here is the distribution of X:
Q1: What is the probability that the
student got a B or better?
Ans: P(X ≥ 3 ) = P(X=3) + P(X=4)
= 0.40 + 0.31 = 0.71
Q2: Suppose that a grade of D or F in English210 will not count as satisfying
a requirement for a major in linguistics. What is the probability that a
randomly selected student will not satisfy this requirement?
Ans: P(X ≤ 1 ) = 1 - P( X >1) = 1 – ( P(X=2) + P(X=3) + P(X=4) ) = 1- 0.91 = 0.09

8
Continuous Random Variable
A continuous random variable Y takes on all values in an interval of
numbers.
Ex: Suppose we want to choose a number at random between 0 and 1.
-----There is infinitely many number between 0 and 1.
How do we assign probabilities to events in an infinite sample space?
 The probability distribution of Y is described by a density curve.
 The probability of any event is the area under the density curve and
above the values of Y that make up the event.

9
 A discrete random variable X has a finite number of possible values.
The probability model of a discrete random variable X assigns a
probability between 0 and 1 to each possible value of X.
 A continuous random variable Y has infinitely many possible values.
The probability of a single event (ex: X=k) is meaningless for a
continuous random variable. Only intervals can have a non-zero
probability; represented by the area under the density curve for that
interval .
 Discrete random variables commonly arise from situations that
involve counting something.
 Situations that involve measuring something often result in a
continuous random variable.
Continuous Random Variable (Cont…)

10
Continuous Probability Models
Example: This is a uniform density curve for the variable X. Find the
probability that X falls between 0.3 and 0.7.
Ans: P(0.3 ≤ X ≤ 0.7) = (0.7- 0.3) * 1 = 0.4
Uniform
Distribution

11
Continuous Probability Models (Cont…)
Example: Find the probability of getting a random number that is
less than or equal to 0.5 OR greater than 0.8.
P(X ≤ 0.5 or X > 0.8)
= P(X ≤ 0.5) + P(X > 0.8)
= 0.5 + 0.2
= 0.7
Uniform
Distribution

12
Continuous Probability Models (Cont…)
General Form:
The probability of the event A is the shaded area under the density
curve. The total area under any density curve is 1.

13
Normal Probability Model
The probability distribution of many random variables is a normal
distribution.
Example: Probability distribution
of Women’s height.
Here, since we chose a woman
randomly, her height, X, is a
random variable.
To calculate probabilities with the normal distribution, we standardize
the random variable (z score) and use the Table A.

14
Normal Probability Model (Cont…)
Reminder: standardizing N(µ,σ)
We standardize normal data by calculating z-score so that any normal
curve can be transformed into the standard Normal curve N(0,1).
σ
µ)( −
=
x
z

15
Normal Probability
Model (Cont…)
Women’s heights are normally
distributed with µ = 64.5 and σ = 2.5
in.
The z-scores for 68,
And for x = 70",
4.1
5.2
)5.6468(
=
−
=z
z =
(70−64.5)
2.5
= 2.2
The area under the curve for the interval
[68”,70”] is 0.9861-0.9192=0.0669.
Thus the probability that a randomly
chosen woman falls into this range is
6.69%. i.e.
P(68 ≤ X ≤ 70)= 6.69%.
What is the probability, if we pick one woman at random, that her height
will be between 68 and 70 inches i.e. P(68 ≤ X ≤ 70)? Here because the
woman is selected at random, X is a random variable.

Chapter 4 part2- Random Variables

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Chapter 4 part2- Random Variables