Section 6.1: Random Variables
Random Variables (r.v.): a variable whose value is a numerical outcome of a random phenomenon.
*denoted with capital letters. Usually X & Y.
Discrete Random Variable: has a countable number of possible outcomes.
k outcomes
Probability Distribution:
*In general:
Value of X
Probability
X1
p1
…
…
X2
p2
Xk
pk
Recall: 1. 0 ≤ pi ≤ 1 (all probabilities have to be between 0 and 1)
2. p1  p2  ...  pk  1 (all probabilities in a sample space add to 1)
Example, p. 349
Apgar Scores
X = Apgar score of a randomly selected baby one minute after birth
Value:
Probability:
0
0.001
1
0.006
2
0.007
3
0.008
4
0.012
5
0.020
6
0.038
7
0.099
8
0.319
9
0.437
10
0.053
(a) Show that this probability distribution for X is legitimate.
 Each probability is between 0 and 1.
 0.001 + 0.006 + 0.007 + 0.008 + 0.012 + 0.020 + 0.038 + 0.099 + 0.319 + 0.437 + 0.053 = 1.000
(b) Make a histogram of the probability distribution.
Probability Histogram:
* remember columns equal width
* columns touch
* probability will be the height
(c) Doctors decided that Apgar scores of 7 or higher indicate a healthy baby. What’s the probability that a
randomly selected baby is healthy?
P(X ≥ 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.099 + 0.319 + 0.437 + 0.053 = 0.908
We have about a 91% chance of randomly choosing a healthy baby.
Mean of Discrete Random Variables (Expected Value)
Value of X
Probability
x1
p1
…
…
x2
p2
 X  x1 p1  x2 p2  ...  xk pk   xi pi
xk
pk
On AP Formula chart
Variance: How spread out the data is.
Remember s2  Standard deviation s. This was for a sample. We will now be using σ2.
 2  ( x1  x )2 p1  ( x2  x )2 p2  ...  ( xk  x )2 pk   ( xi  x )2 pi
Examples, p. 352 & 353
Apgar Scores
Value:
Probability:
0
0.001
1
0.006
2
0.007
3
0.008
4
0.012
5
0.020
6
0.038
7
0.099
8
0.319
9
0.437
10
0.053
Compute the mean and standard deviation of the random variable X. Interpret each in context.
*** 𝜇𝑥 = 0(0.001) + 1(0.006) + 2(0.007) + ⋯ + 9(0.437) + 10(0.053) = 8.128  Avg. Apgar Score or
Expected Score
Interpret: If many, many babies were randomly selected, the average Apgar score would be 8.128.
***𝜎𝑥 2 = (0 − 8.128)2 (0.001) + (1 − 8.128)2 (0.006) + ⋯ + (9 − 8.128)2 (0.437) + (10 −
8.128)2 (0.053) ≈ 2.066
To calculate the standard deviation: √𝜎𝑥 2 = √2.066 ≈ 1.437
Interpret: A randomly selected baby’s Apgar score will typically differ from the mean by about 1.4 units.
Continuous Random Variable: X takes all values in an interval of numbers. The probabilitiy distribution is
described by a density curve.
0
Imagine a spinner:
* probability of getting one exact value is 0
* for cont. r.v.s, we’re interested in intervals.
1/4
3/4
1/2
S = {all numbers x such that 0 ≤ x ≤ 1}
S = {X: 0 ≤ x ≤ 1}
No probability distribution!
We will use a uniform density curve.
Total Area = 1
1
P(0.3 ≤ X ≤ 0.7)
Find the area!
1 ∙ (0.7 – 0.3) = 1 ∙ 0.4 = 0.4
0
0.3
0.7
1
***With continuous random variables, the signs don’t matter (< or ≤ give you the same answer).
Normal distributions as probability distributions
Recall: N(µ, σ)  Normal distribution with mean µ and standard deviation σ
To standardize X if it has N(µ, σ): z 
X 

 Changes it to N(0, 1)
Example, p. 357
Young Women’s Heights: N(64, 2.7)  Normally distributed with µ = 64 inches and σ = 2.7 inches.
Y = Height of randomly selected young woman.
What’s the probability that a randomly chosen young woman has a height between 68 and 70 inches?
P(68 ≤ Y ≤ 70) =
P(
68−64
2.7
≤𝑧≤
70−64
2.7
)
= P (1.48 ≤ 𝑧 ≤ 2.22)
= 0.9868 – 0.9306 = 0.0562
There’s about a 5.6% chance that a randomly chosen young woman will be between 68 and 70 inches tall.
Homework: p. 359 # 19 b & c, 21, 24, 27, 28, 30