1. 6.1 Random Variables
Random variable – variable whose value is a numerical outcome of a random phenomenon.
Discrete – countable number of number of probable outcomes
Continuous – measurable ex. Temperature, height, age

2. Discrete variables Discrete example Find P(x=0) Find P(x≤2)
Cars owned x
Number of families
relative frequency or P(x)
xP(x) 17 1
877
0.219 2
1767 3
1107
0.83 4
200
0.05 5 32
0.008
4000

3. Random variables Population mean μ also called “expected value”
μ = Σ xP(x)
variance
σ2 = Σ (x- μ)2P(x)
Standard deviation = σ
Find μ and σ for last problem

4. Continuous variables
What was the age distribution of nurses in Great Britain at the time of Florence Nightingale nurses in Great Britain?
Age
Midpoint
P(x)
xP(x)
x-μ
(x-μ)2
(x-μ)2P(x)
20-29
24.5
0.057
1.4
-29.3
856
48.8
30-39
34.5
0.097
3.35
-19.3
371 36
40-49
44.5
0.197
8.77
-9.26
85.7
16.9
50-59
54.5
0.292
15.9
0.74
0.55
0.16
60-69
64.5
0.25
16.1
10.7
115
28.8
70-79
74.5
0.091
6.78
20.7
430
39.1
80+
84.5
0.018
1.52
30.7
945 17
1.002
53.8
187
13.7

5. Continuous variables Draw a histogram of the distribution
Find the probability that a nurse selected at random would be 60 or older.
What is the expected age of nurses in 1851?
Find the standard deviation.

6. Random variables
USA Today reported that about 25% of all state prisoners released on Parole become repeat offenders while on parole. Suppose the parole board is examining 5 prisoners for parole. Let x= number of prisoners of the 5 on parole who become repeat offenders.
Here is the probability distribution:

7. Random variables x 1 2 3 4 5 P(x) 0.24 0.4 0.26 0.09 0.021 2 3 4 5
P(x)
0.24
0.4
0.26
0.09
0.02
Find the probability that one or more of the 5 will be repeat offenders?
Find the probability that 2 or more will be repeat offenders?
Find the probability that 4 or more will be repeat offenders?
Find the expected number of repeat offenders?
Find the standard deviation.