Chapter 7. 7 multiple choice and 8 worked questions, all with solutions.
These are the Checkpoint 2 practice questions for this topic. Work them with the answers covered.
1. A help desk logs 0, 1, 2 or 3 escalations an hour with probabilities 0.15, 0.30, 0.35, 0.20. The expected number per hour is
(A) 1.30 (B) 2.00 (C) 1.60 (D) 1.50
2. For that same help desk, the variance of escalations per hour is
(A) 3.5000 (B) 2.5600 (C) 1.4400 (D) 0.9400
3. Each escalation costs $90 to resolve and the desk carries $200 of fixed hourly cost. Expected hourly cost is
(A) 344.00 (B) 144.00 (C) 434.00 (D) 201.60
4. For that same cost, the standard deviation is closest to
(A) 7614.00 (B) 0.97 (C) 84.60 (D) 87.26
5. A courier’s daily profit has mean $400 and variance 900. If a $50 fixed fee is added to every day’s profit, the new variance is
(A) 950 (B) 900 (C) 2500 (D) 1000
6. Two warehouses have daily cost variances 25 and 16, with covariance $-10$. The variance of their combined daily cost is
(A) 21 (B) 41 (C) 61 (D) 31
7. Two shops have sales variances 9 and 25, correlation 0.4. Their covariance is
(A) 3.6 (B) 6 (C) 90 (D) 0.4
1. A machine produces 0, 1, 2 or 3 faults per shift with probabilities 0.15, 0.30, 0.35, 0.20. (a) Verify the distribution is valid. (b) Find $E(X)$. (c) Find $V(X)$ and $\sigma$.
2. Each fault costs $80 to clear and the shift carries $250 of fixed cost, so cost is $C = 80X + 250$ for the $X$ above. Find $E(C)$, $V(C)$ and $\sigma_C$.
3. $E(X) = 20$ and $E(X^2) = 449$. Find $V(X)$ and $\sigma$, then find $V(2X+7)$.
4. Two product lines have $V(X) = 36$, $V(Y) = 64$ and correlation $0.25$. (a) Find the covariance. (b) Find $V(X+Y)$. (c) What would $V(X+Y)$ be if the lines were uncorrelated, and what is the lesson?
5. A distribution has $\mathrm{P}(X{=}1){=}0.4$, $\mathrm{P}(X{=}2){=}0.35$ and $\mathrm{P}(X{=}3){=}k$. Find $k$, then $E(X)$ and $V(X)$.
6. Explain why the expected value of a random variable need not be a value the variable can take, giving an example.
7. Derive $V(X) = E(X^2) - \mu^2$ from the definition $V(X) = E[(X-\mu)^2]$.
8. $X$ has $E(X) = 5$, $V(X) = 9$. Find $E(X^2)$.
1. C — A weighted sum. 1.50 is the unweighted average of 0,1,2,3, which ignores how likely each is.
2. D — $E(X^2) = 3.50$, so $V(X) = 3.50 - 1.60^2$. The distractors are $E(X^2)$ and $[E(X)]^2$ reported on their own.
3. A — $90E(X) + 200$. The first distractor drops the fixed cost.
| 4. D — $ | a | \sigma = 90\sqrt{V(X)}$. The first distractor is the variance, un-rooted. |
5. B — Adding a constant shifts the distribution without stretching it, so $V(X+b)=V(X)$.
6. A — $25+16+2(-10)$. 41 omits the covariance entirely.
7. B — $\rho\sigma_x\sigma_y = 0.4(3)(5)$. 3.6 multiplies by the variances rather than the standard deviations.
1. (a) All four lie in $[0,1]$ and sum to 1.00. (b) $E(X) = 1.60$. (c) $E(X^2) = 3.50$, so $V(X) = 0.9400$ and $\sigma = 0.9695$.
2. $E(C) = 80(1.60) + 250 = 378.00$. $V(C) = 80^2(0.9400) = 6016.00$, so $\sigma_C = 77.56$. The $+250$ shifts the mean and leaves the variance untouched.
3. $V(X) = 449 - 400 = 49$, so $\sigma = 7$. And $V(2X+7) = 4(49) = 196$.
4. (a) $0.25(6)(8) = 12$. (b) $36+64+2(12) = 124$. (c) It would be 100. Positive correlation adds 24 to the variance of the total for no extra return, which is the case for diversifying into things that do not move together.
5. $k = 0.25$. $E(X) = 0.4+0.7+0.75 = 1.85$. $E(X^2) = 0.4+1.4+2.25 = 4.05$, so $V(X) = 4.05 - 3.4225 = 0.6275$.
6. $E(X)$ is a balance point, a probability-weighted average, not an outcome. A die has $E(X) = 3.5$, which no face shows; a household can average 2.4 children.
7. Expand: $E[X^2 - 2\mu X + \mu^2] = E(X^2) - 2\mu E(X) + \mu^2$, using that expectation is linear and $\mu$ is a constant. Since $E(X) = \mu$, this is $E(X^2) - 2\mu^2 + \mu^2 = E(X^2) - \mu^2$.
8. Rearranging the shortcut, $E(X^2) = V(X) + \mu^2 = 9 + 25 = 34$.