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.
The answer is 3.667 for part (c). Our formula begins 1(1/15) + 2(2/15) + 

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 substitute each value of x (1, 2, 3, 4, 5) into the formula P(x) = x/15:

P(1) = 1/15
P(2) = 2/15
P(3) = 3/15
P(4) = 4/15
P(5) = 5/15

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 if:
1. Every probability is between 0 and 1: All the probabilities in the distribution are fractions, which are between 0 and 1.
2. The sum of all probabilities is equal to 1: We can add up all the probabilities in the distribution:

P(1) + P(2) + P(3) + P(4) + P(5) = (1/15) + (2/15) + (3/15) + (4/15) + (5/15) = (15/15) = 1

Since the sum is equal to 1, the probability distribution of x satisfies the properties of a discrete probability distribution.

c) To calculate the mean of x, we multiply each value of x by its corresponding probability and sum them up:

Mean (µ) = 1 * (1/15) + 2 * (2/15) + 3 * (3/15) + 4 * (4/15) + 5 * (5/15)

Simplifying the equation, we get:

µ = (1 + 4 + 9 + 16 + 25)/15

µ = 55/15

µ ≈ 3.667

Therefore, the mean of x is approximately 3.667.

d) To calculate the variance and standard deviation, we need to use the formula:

Variance (σ^2) = ∑[(x - µ)^2 * P(x)]
Standard Deviation (σ) = √Variance

For each value of x, we subtract the mean (µ), square the result, multiply it by the corresponding probability (P(x)), and sum them up. Then, take the square root of the variance to calculate the standard deviation.

Variance (σ^2) = (1 - 3.667)^2 * (1/15) + (2 - 3.667)^2 * (2/15) + (3 - 3.667)^2 * (3/15) + (4 - 3.667)^2 * (4/15) + (5 - 3.667)^2 * (5/15)

Simplifying the equation, we get:

σ^2 = (6.111 + 1.111 + 0.111 + 0.111 + 1.111)/15

σ^2 = 8.555/15

σ^2 ≈ 0.570

Standard Deviation (σ) = √0.570 ≈ 0.755

Therefore, the variance is approximately 0.570 and the standard deviation is approximately 0.755.

For the second part of the question regarding normal curves, it seems unrelated to the given context. Could you please provide more information or clarify your question?