Question

A pizza shop delivers an extra large pizza, which is sold for \$20$20dollar sign, 20, and costs the pizza shop \$12$12dollar sign, 12 to make. The pizza shop has a delivery policy that says if the pizza takes longer than half an hour to arrive, there is no charge. Experience has shown that delivery takes longer than half an hour only 10\%10%10, percent of the time.
Let the random variable XXX be the pizza shop's profit for a randomly selected delivery order for one of these pizzas. Here is the probability distribution for XXX:
Late On-time
X=\text{profit}X=profitX, equals, start text, p, r, o, f, i, t, end text -\$12−$12minus, dollar sign, 12 \$8$8dollar sign, 8
P(X)P(X)P, left parenthesis, X, right parenthesis 0.100.100, point, 10 0.900.900, point, 90
Given that \mu_X=\$6μ
X
​
=$6mu, start subscript, X, end subscript, equals, dollar sign, 6, calculate \sigma_Xσ
X
​
sigma, start subscript, X, end subscript.
Round your answer to two decimal places.
\sigma_X=σ
X
​
=sigma, start subscript, X, end subscript, equals
dollars

Answer 1

The standard deviation for the given probability distribution is;

σₓ = $6

Probability Distribution

The image for the probability distribution is missing and so i have attached it.

We are given;

Expected Value which is the mean as;

µₓ = $6

Now, from the probability distribution table attached, we can find the mean (expected value) of a discrete random variable X from;

Variance(X) = (-12 - 6)²(0.1) + (8 - 6)²(0.9) = 36

Thus, standard deviation of the random X is;

σₓ = √36

σₓ = $6

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