fro x = 12, 1, 32, 2 if the set of x is different just follow the same procedure for your particular set.
Answer:
c = 1/1173
E[x] = 29.416
Step-by-step explanation:
The probability mass function is the probability that a discrete random variable X takes a particular value of x for the probability mass function the sum of all the items in the mass function should be equal one, therefore, 1 = c(12^2 + 1^2 + 32^2 + 2^2) = 1, solving for c we get c = 1/1173, our probability mass function is p(x) = 1/1172*x^2 for x = 12, 1, 32, 2
The expected value is the probability-weighted average of all possible values, to calculate it we need to sum each item x by its respective probability:
E[X] = 12*(1/1173*12^2)+1*(1/1173*1^2)+32(1/1173*32^2)+2(1/1173*2^2) = 29.416
Please note that to calculate E[X] we need to calculate c first to have the probabilities of each x
(a) The value of [tex]c=\dfrac{1}{1173}[/tex]
(b) The expected value will be [tex]X=29.416[/tex]
The probability mass function is the probability that a discrete random variable X takes a particular value of x for the probability mass function.
For the mass function, the sum of all the given items should be equal to one.
Therefore,
[tex]c(12^2+1^2+32^21+2^2)=1[/tex]
By solving for c we get
[tex]c=\dfrac{1}{1173}[/tex]
The probability mass function is given as
[tex]P(x)=\dfrac{1}{1172} \times x^2[/tex]
for x = 12, 1, 32, 2
To find out the expected value of probability we will put every value of x in the above formula and the sum will be the expected probability.
[tex]X=(12\times (\dfrac{1}{1173})\times12^2)+ (1\times (\dfrac{1}{1173})\times1^2)+(32\times (\dfrac{1}{1173})\times32^2)+(2\times (\dfrac{1}{1173})\times2^2)=29.416[/tex]
[tex]X=29.416[/tex]
Thus
(a) The value of [tex]c=\dfrac{1}{1173}[/tex]
(b) The expected value will be [tex]X=29.416[/tex]
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