The following table shows the probability distribution for a discrete random variable. X 11 14 16 19 21 23 24 29 P(X) 0.07 0.21
2 answers:
Because the random variable X is discrete, the variance of X is defined as
var(X) =
where
p = values of probability as given in the table.
μ = 18.59
Calculate the array (x - μ).
x-μ = [-7.59 -4.59 -2.59 0.41 2.41 4.41 5.41 10.41]
Calculate the array (x-μ)².
(x- μ)² = [57.6081 21.0681 6.7081 0.1681 5.8081 19.4481 29.2681 108.3681]
Calculate the array p*(x-μ)².
p*(x- μ)² = [4.0326 4.4342 1.1404 0.042 0.2904 0.7779 3.8049 8.6694]
Calculate the variance of X.
var(X) = 23.182
Answer: 23.18 (nearest hundredth)
var(X) =
where
p = values of probability as given in the table.
μ = 18.59
Calculate the array (x - μ).
x-μ = [-7.59 -4.59 -2.59 0.41 2.41 4.41 5.41 10.41]
Calculate the array (x-μ)².
(x- μ)² = [57.6081 21.0681 6.7081 0.1681 5.8081 19.4481 29.2681 108.3681]
Calculate the array p*(x-μ)².
p*(x- μ)² = [4.0326 4.4342 1.1404 0.042 0.2904 0.7779 3.8049 8.6694]
Calculate the variance of X.
var(X) = 23.182
Answer: 23.18 (nearest hundredth)
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Answer:
a) 0.4219 = 42.19% probability that one out of the next four cars needs oil.
b) 0.3115 = 31.15% probability that two out of the next eight cars needs oil.
c) 0.2581 = 25.81% probability that three out of the next 12 cars need oil.
Step-by-step explanation:
For each car, there are only two possible outcomes. Either they need oil, or they do not need it. The probability of a car needing oil is independent of any other car, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
One out of four cars needs to have oil added.
This means that
a. One out of the next four cars needs oil.
This is P(X = 1) when n = 4. So
0.4219 = 42.19% probability that one out of the next four cars needs oil.
b. Two out of the next eight cars needs oil.
This is P(X = 2) when n = 8. So
0.3115 = 31.15% probability that two out of the next eight cars needs oil.
c.Three out of the next 12 cars need oil.
This is P(X = 3) when n = 12. So
0.2581 = 25.81% probability that three out of the next 12 cars need oil.