36 possibilities
number of times sum
1 * 2 = 2
2 * 3 = 6
3 * 4 = 12
4 * 5 = 20
5 * 6 = 30
6 * 7 = 42
5 * 8 = 40
4 * 9 = 36
3 * 10= 30
2 * 11= 22
1 * 12= 12
-----------
36 and 252
mean = 252/36 = 7
go ahead and compute sigma
sample space is defined to be the sum of two numbers turning up on the toss. Find the
discrete probability distribution for the random variable X and compute the corresponding
mean and standard deviation.
number of times sum
1 * 2 = 2
2 * 3 = 6
3 * 4 = 12
4 * 5 = 20
5 * 6 = 30
6 * 7 = 42
5 * 8 = 40
4 * 9 = 36
3 * 10= 30
2 * 11= 22
1 * 12= 12
-----------
36 and 252
mean = 252/36 = 7
go ahead and compute sigma
Sample space (S): {1, 2, 3, 4, 5, 6} for each toss
Now, we can find all possible outcomes of X, which is the sum of two numbers turning up on the toss:
When two dice show 1:
X = 1 + 1 = 2
When two dice show 2:
X = 1 + 2 = 3
X = 2 + 1 = 3
When two dice show 3:
X = 1 + 3 = 4
X = 2 + 2 = 4
X = 3 + 1 = 4
When two dice show 4:
X = 1 + 4 = 5
X = 2 + 3 = 5
X = 3 + 2 = 5
X = 4 + 1 = 5
When two dice show 5:
X = 1 + 5 = 6
X = 2 + 4 = 6
X = 3 + 3 = 6
X = 4 + 2 = 6
X = 5 + 1 = 6
When two dice show 6:
X = 1 + 6 = 7
X = 2 + 5 = 7
X = 3 + 4 = 7
X = 4 + 3 = 7
X = 5 + 2 = 7
X = 6 + 1 = 7
Now, let's calculate the probabilities for each outcome:
P(X = 2) = 1/36
P(X = 3) = 2/36 = 1/18
P(X = 4) = 3/36 = 1/12
P(X = 5) = 4/36 = 1/9
P(X = 6) = 5/36
P(X = 7) = 6/36 = 1/6
To calculate mean (μ) and standard deviation (σ), we can use the following formulas:
Mean (μ) = ∑X * P(X)
Standard Deviation (σ) = √∑(X-μ)² * P(X)
Let's calculate:
Mean (μ) = 2 * (1/36) + 3 * (1/18) + 4 * (1/12) + 5 * (1/9) + 6 * (5/36) + 7 * (1/6)
= (2 + 3/2 + 4/3 + 5/2 + 10/3 + 7/6)
= (12/6 + 9/6 + 8/6 + 15/6 + 20/6 + 7/6)
= 71/6
≈ 11.83
Standard Deviation (σ) = √((2-μ)² * (1/36) + (3-μ)² * (1/18) + (4-μ)² * (1/12) + (5-μ)² * (1/9) + (6-μ)² * (5/36) + (7-μ)² * (1/6))
First, substitute the value of μ = 71/6 into the formula.
Standard Deviation (σ) = √((2 - 71/6)² * (1/36) + (3 - 71/6)² * (1/18) + (4 - 71/6)² * (1/12) + (5 - 71/6)² * (1/9) + (6 - 71/6)² * (5/36) + (7 - 71/6)² * (1/6))
Now, calculate the standard deviation (σ) using the above formula. It will be approximately equal to a specific value.
Let's start by listing all the possible outcomes when a die is tossed twice:
1 + 1 = 2, 1 + 2 = 3, 1 + 3 = 4, 1 + 4 = 5, 1 + 5 = 6, 1 + 6 = 7
2 + 1 = 3, 2 + 2 = 4, 2 + 3 = 5, 2 + 4 = 6, 2 + 5 = 7, 2 + 6 = 8
3 + 1 = 4, 3 + 2 = 5, 3 + 3 = 6, 3 + 4 = 7, 3 + 5 = 8, 3 + 6 = 9
4 + 1 = 5, 4 + 2 = 6, 4 + 3 = 7, 4 + 4 = 8, 4 + 5 = 9, 4 + 6 = 10
5 + 1 = 6, 5 + 2 = 7, 5 + 3 = 8, 5 + 4 = 9, 5 + 5 = 10, 5 + 6 = 11
6 + 1 = 7, 6 + 2 = 8, 6 + 3 = 9, 6 + 4 = 10, 6 + 5 = 11, 6 + 6 = 12
Since each outcome has an equal probability of 1/36, we can determine the probability for each sum by counting the number of outcomes that result in that sum.
The probability distribution for X is as follows:
X = 2, P(X = 2) = 1/36
X = 3, P(X = 3) = 2/36
X = 4, P(X = 4) = 3/36
X = 5, P(X = 5) = 4/36
X = 6, P(X = 6) = 5/36
X = 7, P(X = 7) = 6/36
X = 8, P(X = 8) = 5/36
X = 9, P(X = 9) = 4/36
X = 10, P(X = 10) = 3/36
X = 11, P(X = 11) = 2/36
X = 12, P(X = 12) = 1/36
To compute the corresponding mean, we can use the formula:
Mean (μ) = ∑ (x * P(X = x)), where x represents the possible values of X.
Mean (μ) = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + (5 * 4/36) + (6 * 5/36) + (7 * 6/36) + (8 * 5/36) + (9 * 4/36) + (10 * 3/36) + (11 * 2/36) + (12 * 1/36)
Simplifying the equation gives:
Mean (μ) = 7
The standard deviation (σ) can be calculated using the formula:
Standard Deviation (σ) = √(∑ ((x - μ)^2 * P(X = x)))
Standard Deviation (σ) = √(((2 - 7)^2 * 1/36) + ((3 - 7)^2 * 2/36) + ((4 - 7)^2 * 3/36) + ((5 - 7)^2 * 4/36) + ((6 - 7)^2 * 5/36) + ((7 - 7)^2 * 6/36) + ((8 - 7)^2 * 5/36) + ((9 - 7)^2 * 4/36) + ((10 - 7)^2 * 3/36) + ((11 - 7)^2 * 2/36) + ((12 - 7)^2 * 1/36))
Simplifying the equation gives:
Standard Deviation (σ) ≈ 2.415
Therefore, the mean of the random variable X is 7 and the standard deviation is approximately 2.415.