Suppose that the probability distribution of a random variable x can be described by the formula
P(x) = x/15
For each of the values x = 1, 2, 3, 4, and 5. For examples, then, p(x=2) = p(2) = 2/15
a Write out the probability distribution of x.
b Show that the probability distribution of x satisfies the properties of a discrete probability distribution.
c Calculate the mean of x.
d Calculate the variance and standard deviation.
In each case, sketch the two specified normal curves on the same set of axes:
A normal curve with µ = 20 and o = 3, a normal curve with µ = 20 and o = 6.
a) To write out the probability distribution of x, we can substitute the given values of x into the formula P(x) = x/15:
P(x=1) = 1/15
P(x=2) = 2/15
P(x=3) = 3/15
P(x=4) = 4/15
P(x=5) = 5/15
So, the probability distribution of x is:
P(x) = {1/15, 2/15, 3/15, 4/15, 5/15}
b) To show that the probability distribution of x satisfies the properties of a discrete probability distribution, we need to check two main properties:
1. The sum of the probabilities should be equal to 1:
Sum(P(x)) = (1/15) + (2/15) + (3/15) + (4/15) + (5/15) = 15/15 = 1
2. The probabilities should be non-negative:
P(x) is always equal to x/15, and since x takes on values 1, 2, 3, 4, and 5, the probabilities are positive and satisfy this condition.
Therefore, the probability distribution of x satisfies the properties of a discrete probability distribution.
c) To calculate the mean of x, we need to find the expected value, denoted as E(x), which is the weighted average of the values of x:
E(x) = (1/15) * 1 + (2/15) * 2 + (3/15) * 3 + (4/15) * 4 + (5/15) * 5
E(x) = (1 + 4 + 9 + 16 + 25) / 15 = 55/15 = 3.67
So, the mean of x is 3.67.
d) To calculate the variance and standard deviation of x, we need to use the following formulas:
Variance (Var(x)) = E(x^2) - [E(x)]^2
Standard Deviation (SD(x)) = sqrt(Var(x))
To calculate E(x^2), we need to find the expected value of x^2:
E(x^2) = (1/15) * 1^2 + (2/15) * 2^2 + (3/15) * 3^2 + (4/15) * 4^2 + (5/15) * 5^2
E(x^2) = (1 + 8 + 27 + 64 + 125) / 15 = 225/15 = 15
Now, we can calculate the variance:
Var(x) = E(x^2) - [E(x)]^2
Var(x) = 15 - (3.67)^2
Var(x) ≈ 1.39
Finally, we can calculate the standard deviation:
SD(x) = sqrt(Var(x))
SD(x) ≈ sqrt(1.39) ≈ 1.18
So, the variance of x is approximately 1.39 and the standard deviation is approximately 1.18.
Regarding the second part of your question about normal curves, it seems unrelated to the previous probability distribution problem. Could you please clarify or provide more information?