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



1/7+2x/3=(15x-3)/21 make all terms have a common denominator of 21
3+14x=15x-3 subtract 3 from both sides
14x=15x-6 subtract 15x from both sides
-x=-6 divide both sides by -1
x=6
*Notice the use of parentheses in this format to make it clear what the numerators and denominators are...
B would be 5.40 because inches and yards would be 135
Answer:
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Answer:
y-1=3(x--1)= 3x-y+4
Step-by-step explanation:
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