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



Answer:
u = (b/hn)
Step-by-step explanation:
The inverse!!
You would multiply 62 1/4 and 4. Which you cant do that so you would have to make it into a inproper fraction so that will be 249/4 times 4/1 then you can multiply which your answer would be 996 /4 and then in simplest form would be 249/1= 249
Answer:
y=63
x=63
z=72
Step-by-step explanation:
Part A: 40/100 = 0.4 0.4*36=14 14 + 36=50 Part B :40/100=0.4 0.4*60=24 60+24=84
hope this helped