Answer:
We can find the second moment given by:
And we can calculate the variance with this formula:
And the deviation is:
Step-by-step explanation:
For this case we have the following probability distribution given:
X 0 1 2 3 4 5
P(X) 0.031 0.156 0.313 0.313 0.156 0.031
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
We can verify that:
And
So then we have a probability distribution
We can calculate the expected value with the following formula:
We can find the second moment given by:
And we can calculate the variance with this formula:
And the deviation is:
Answer:
C
Step-by-step explanation:
The multiples of 3 are
3, 6, 9, 12, 15, ...........
The terms form an arithmetic sequence with common difference of 3.
The sum to n terms is
= [2a₁ + (n - 1)d ]
where a₁ is the first term and d the common difference
Here a₁ = 3 and d = 3, thus
= [ (2 × 3) + (139 × 3) ]
= 70(6 + 417) ] = 70 × 423 = 29610 → C
33/5*4/9=11*3/5 * 2*2/3*3 = 44/15 = 2.93333333
A single value randomly selected from the distribution of W will vary 4 by 3.5 attempted experiments. (collegboard)
$431.85. I got this by multiplying the amount of weeks in a year (52) by two to get 104. I then multiplied that by 3.5 to get 364. I then added 32.85 and 35 to that to get $431.85