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Suppose that the expectation of a random variable $X$ is $5$. Which of the following statements is true?

1. There is a sample point at which $X$ has the value $5$.
2. There is a sample point at which $X$ has value greater than $5$.
3. There is a sample point at which $X$ has a value greater than equal to $5$.
4. None of the above.

For this graph random variable X is greater than 5 . But if a graph is negatively skewed or symmetrical then value can be less than and equal to 5 respectively . So value can be greater than , less than  or equal to 5 .
This is a Simple Diagram or generalized Diagarm

It has no relation with any Point.

(Bolded one is probability p and non-bolded is random variable x so written of form p*x) ---

E(x) =  0.8*6 + 0.2*1 = 5   (Option- A)

E(x) = 0.8*5  +  0.2*5  = 5 (Option – B)

Expectation of discrete random variable (finite case)

$E(X) = x_1 p_1 + x_2p_2 + \dots +x_np_n$

$E(X) = 5, 0 \leq pi \leq1$

$p_1 + p_2 + \dots + p_n = 1$

Therefore, $E(X) = 5$ is possible only if at-least one of the  $x_i \geq 5$

by

First of all option B is not true. But you're interpreting it wrongly. I would suggest to do Mathematical logic first before doing any other subject.
"has a value greater than equal to 5" does not say that all values are greater than 5.
Thanks. Now I got it

Expectation is nothing but average value ...

A) Let the samples be 4,6 .. Average = (4+6) / 2 = 5 .. So average of 5 is possible even if no sample point has a value 5... So A) is eliminated.

B) Let samples be 5,5 .. Average = (5+5) / 2 = 5 .. So Average can be 5 even if no sample is greater than 5 .. So B) is eliminated ...

C) should be always TRUE .. If you dont have even 1 sample which is <= average_value x then average of x is not possible.. This is also applicable if you dont have even 1 sample point which is >= average_value

So Option C) is the answer ...

It's weighted average.
yes

@Aakash_ mean and average are calculated on sample data . but expectation is calculated on real probability . Its a theoretical concept.

I think answer should be c) as I could not find any case in which mean is 5 and at all sample points X has value less than 5
by

For example if X = {0,1,2,3,4}

and if P1=P2=P3=P4=P5 = ½

So E(X) = 0(1/2) + 1(1/2) + 2(1/2) + 3(1/2) + 4(1/2)  = 10(1/2) = 5. In this case none of the options are satisfied. So why is (d) not the correct answer ?
Sum of probabilities p1, P2,....p5 should be equal to 1