Question

Let x represent the # of times a student visits a nearby pizza shop in one month period. Assume the following table is the distribution of x. Ie x =0 and p(x) is 0.10, c=1 for a p(x) of 0..30, x=2 for a p(x)=0.40 and x=3 for a p(x) of 0.20 with x = to the # of times a student visits the pizza parlor. What are the expected value and the standard deviation of x? Be sure to show your work.

If I were to compute the conventional mean of X, my answer would be X bar =1.5. Why does the answer in A ) differ?

Answer 1

To find the expected value and standard deviation of the distribution of x, we need to calculate the weighted average of the values and their corresponding probabilities. Let's work through the calculations:

Given:
x = 0, p(x) = 0.10
x = 1, p(x) = 0.30
x = 2, p(x) = 0.40
x = 3, p(x) = 0.20

To find the expected value (E[x]), we multiply each value of x by its corresponding probability and sum them up:

E[x] = (0 * 0.10) + (1 * 0.30) + (2 * 0.40) + (3 * 0.20)
= 0 + 0.30 + 0.80 + 0.60
= 1.70

Therefore, the expected value of x is 1.70.

Now, let's calculate the standard deviation (σ) of x. We'll use the formula:

σ = sqrt(E[(x - E[x])^2])

First, calculate (x - E[x])^2 for each value of x, then multiply it by its corresponding probability and sum them up:

[(0 - 1.70)^2 * 0.10] + [(1 - 1.70)^2 * 0.30] + [(2 - 1.70)^2 * 0.40] + [(3 - 1.70)^2 * 0.20]
= [(-1.70)^2 * 0.10] + [(-0.70)^2 * 0.30] + [(-0.30)^2 * 0.40] + [(1.30)^2 * 0.20]
= [2.89 * 0.10] + [0.49 * 0.30] + [0.09 * 0.40] + [1.69 * 0.20]
= 0.289 + 0.147 + 0.036 + 0.338
= 0.810

Now take the square root of the result to find the standard deviation:

σ = sqrt(0.810)
= 0.900

Therefore, the standard deviation of x is approximately 0.900.

Regarding the difference in the conventional mean (X bar) of 1.5, it appears to be incorrect based on the given probabilities and values of x. It's important to ensure the calculations consider the correct probabilities and values for each outcome in order to obtain accurate results.