HELP ASAP !! easy question and i’ll give brainliest
2 answers:
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X (adult)= 130
y (child)= 392
(hope you can see it...)
y (child)= 392
(hope you can see it...)
Answer and Explanation:
Given : The random variable x has the following probability distribution.
To find :
a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.
b. Calculate the expected value of x.
c. Calculate the variance of x.
d. Calculate the standard deviation of x.
Solution :
First we create the table as per requirements,
x P(x) xP(x) x² x²P(x)
0 0.25 0 0 0
1 0.20 0.20 1 0.20
2 0.15 0.3 4 0.6
3 0.30 0.9 9 2.7
4 0.10 0.4 16 1.6
∑P(x)=1 ∑xP(x)=1.8 ∑x²P(x)=5.1
a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.
Yes it is a probability distribution.
b) The expected value of x is defined as
c) The variance of x is defined as
d) The standard deviation of x is defined as