The mean of a discrete random variable x is the mean we would expect if the experiment were repeated many times.
The mean value is denoted by μ and obtained with the formula μ = ΣxP (x)
Another term for the mean value of a discrete random variable is expected value.
The expected value is denoted by E (x), i.e. E (x) = ΣxP (x)
In the lesson on the probability distribution of a discrete random variable, we have the following table of probability distribution. Use it to calculate the average number of vehicles owned by people.
|Number of vehicles owned or x||Probability or P (x)|
|P (x) = 1|
How to find the mean of the probability distribution of the number of vehicles people own.
|x||P (x)||xP (x) = x × P (x)|
|0||0.2||0x0.2 = 0|
|1||0.5||1 x 0.5 = 0.5|
|2||0.3||2 x 0.3 = 0.6|
|ΣxP (x) = 0 + 0.5 + 0.6 = 1.1|
E (x) = 1.1
What does an expected value of 1.1 mean for this situation? That means, on average, you would expect people to own around 1.1 vehicles.
Another example showing how to find the mean of a discrete random variable
A survey was carried out to find out how often people go to the cinema each week. After interviewing 500 people, the result is shown in the following table. Let x be the number of visits to the cinema per week. If x = 2, the frequency is 75. This means that 75 people went to the cinema twice a week.
|N = 500|
P (x = 0) = 250/500 = 0.5
P (x = 1) = 125/500 = 0.25
P (x = 2) = 75/500 = 0.15
p (x = 3) = 45/500 = 0.09
P (x = 4) = 5/500 = 0.01
The following table shows the probability distribution.
|P (x) = 1|
How to find the mean for the probability distribution of how often people go to the movies.
E (x) = ΣxP (x) = 0 × 0.5 + 1 × 0.25 + 2 × 0.15 + 3 × 0.09 + 4 × 0.01
E (x) = 0 + 0.25 + 0.30 + 0.27 + 0.04
E (x) = 0.86
Based on the expected value of 0.86, the average number of admissions per week is 0.86.