Download Mean of a Discrete Random Variable

Multiple Choice Practice
1.
A coin is tossed three times. What is the probability that it lands on
heads exactly one time?
(A) 0.125 (B) 0.250
(C) 0.375
(D) 0.500
(E) 0.875
2. Which of the following statements are true?
I. A completely randomized design offers no control for lurking variables.
II. A randomized block design controls for the placebo effect.
III. In a matched pairs design, subjects within each pair receive the same
treatment.
(A) I only (B) II only
(C) III only
(D) All of the above.
(E) None of the above
Solutions
1.
The correct answer is (C). Sample Space: HHH, HHT, HTH, THH, HTT, THT,
TTH, and TTT. Of the eight possible outcomes, three have exactly one
head.
2.
The correct answer is (E). In a completely randomized design , subjects are
randomly assigned to treatment conditions. Randomization provides some
control for lurking variables . By itself, a randomized block design does not
control for the placebo effect . To control for the placebo effect, the
experimenter must include a placebo in one of the treatment levels. In a
matched pairs design , subjects within each pair are assigned to different
treatment levels.
Chap 6.1
Discrete and Continuous
Random Variables
Random Variable
– Takes a numerical value that describes an outcome of a random phenomenon
– Usually represented by capital letters near the end of the alphabet (X, Y, Z)
Probability Distribution of a
Random Variable (X)
– Gives the possible values and their probabilities
Discrete Random Variable
– Has a finitely many possible values.
– All outcomes can be listed
– Each probability must be between 0 and 1
– Sum of the probabilities is 1
Mean of a Discrete Random
Variable
(Expected Value)
– Analyzing discrete random variables: SOCS
– Mean of any discrete random variable is an average of the possible
outcomes, with each outcome weighted by its probability.
Mean of a Discrete Random Variable
(Expected Value)
• To find the mean (or expected value) of “𝑋”,
multiply each possible value by its probability, then
add all the products
• 𝜇𝑋 = 𝑥1 𝑝1 + 𝑥2 𝑝2 + ⋯ + 𝑥𝑘 𝑝𝑘 =
𝑥𝑖 𝑝𝑖
Example
Getting good grades
– Stats 101, fall 2003, grade distribution:
– 21% A’s, 43% B’s, 30% C’s, 5% D’s, 1% F’s
– A student is chosen at random from this class (each student
has the same probability of being chosen)
– Let X = student’s grade (A=4, B=3, C=2, D=1, F=0)
Probability distribution of
Grades:
Probability of getting a B or better?
0.64
Compute the mean of the random variable.
Mean (Expected Value) of a Discrete Random Variable
A baby’s Apgar score is the sum of the ratings on each of five scales,
which gives a whole-number value from 0 to 10.
Let X = Apgar score of a randomly selected newborn
Compute the mean of the random variable X. Interpret this value in
context.
We see that 1 in every 1000 babies would have an Apgar score of 0; 6
in every 1000 babies would have an Apgar score of 1; and so on. So
the mean (expected value) of X is:
The mean Apgar score of a randomly selected newborn is 8.128. This
is the average Apgar score of many, many randomly chosen babies.
Your turn!!
– Make a table for the probability distribution of “Greed”. Let X = the sum
attained when rolling two dice
– Sketch a bar graph to represent the distribution
– Calculate the mean for the random variable and interpret in context.
X
2
P(X)
3
4
5
6
7
8
9
10
11
12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Mean: 7
Continuous Random
Variable
– Takes all values in an interval of numbers
– Described by a density curve
– Cannot list all outcomes
– Probability distributions are described by a density curve
– Assign probability 0 to every individual outcome.
– No distinction between ≥ and >
Normal Distribution
– Are one type of continuous probability distribution.
– Normal curve is most popular density curve
–
N(µ, σ): 𝑍 =
distribution N(0,1)
(𝑋−µ)
is
𝜎
a normal random variable having
Example: Normal probability distributions
The heights of young women closely follow the Normal distribution
with mean µ = 64 inches and standard deviation σ = 2.7 inches.
Now choose one young woman at random. Call her height Y.
If we repeat the random choice very many times, the distribution of
values of Y is the same Normal distribution that describes the heights
of all young women.
Problem: What’s the
probability that the chosen
woman is between 68 and 70
inches tall?
Example: Normal probability distributions
Step 1: State the distribution and the values of interest.
The height Y of a randomly chosen young woman has the
N(64, 2.7) distribution. We want to find P(68 ≤ Y ≤ 70).
Step 2: Perform calculations—
show your work!
The standardized scores for the two
boundary values are
Assignment
6.1 Pg. 359
#3-7 odd, 11, 1415, 21-25 odd, 2730all