Expected Value Decisions Test Answer Key with Scoring – Precalculus

Question 1 Missing Probability (2 points)

(a) P(0) = 1 − (0.25 + 0.30 + 0.15) = 1 − 0.70 = 0.30.

(b) E(X) = (−1)(0.25) + 0(0.30) + 2(0.30) + 5(0.15) = −0.25 + 0 + 0.60 + 0.75 = 1.10.

Answer: (a) 0.30; (b) 1.10

Scoring: 1 point for each part.

Question 2 Reading a Spinner (3 points)

(a) P($0) = 58, P($6) = 28 = 14, P($20) = 18.

(b) E(X) = 0(58) + 6(28) + 20(18) = 128 + 208 = 328 = $4.

(c) Expected net gain = 4 − 5 = −$1, so the game is not fair; the player loses $1 per spin on average.

Answer: E(X) = $4; expected net −$1; not fair

Scoring: 1 point for each part.

Question 3 Cars per Household (2 points)

Divide each count by 25: 0.08, 0.36, 0.40, 0.16. These add to 1.

E = 0(0.08) + 1(0.36) + 2(0.40) + 3(0.16) = 0.36 + 0.80 + 0.48 = 1.64.

Answer: 1.64 cars per household

Scoring: 1 point for the distribution, 1 point for the expected value.

Question 4 Headphone Warranty (3 points)

(a) 0.10(150) = $15.

(b) The warranty costs 48 − 15 = $33 more than the expected repair cost.

(c) On average the warranty is not worth buying. A buyer who could not afford a surprise $150 repair might still choose it to avoid that risk.

Answer: expected repair cost $15 versus $48; do not buy on average

Scoring: 1 point for each part.

Question 5 Insurer’s Expected Profit (3 points)

(a) 0.04(3,500) = $140.

(b) 200 − 140 = $60.

(c) 1,500 × 60 = $90,000.

Answer: $140; $60; $90,000

Scoring: 1 point for each part.

Question 6 A Fair Prize (2 points)

(a) E = 18(16) = $3.

(b) Let the prize be p. A fair game needs p(16) = 6, so p = $36.

Answer: (a) $3; (b) $36

Scoring: 1 point for each part.

Question 7 Comparing Job Offers (3 points)

(a) E = 4,000(0.3) + 2,000(0.5) + 500(0.2) = 1,200 + 1,000 + 100 = $2,300.

(b) Job B’s expected pay is $300 more than Job A’s.

(c) Job B’s pay varies a lot, and a person who needs a dependable income might choose Job A.

Answer: Job B expected pay $2,300 versus $2,000

Scoring: 1 point for each part.

Question 8 Larger of Two Dice (2 points)

Of the 16 equally likely pairs, X = 1 once, X = 2 three times, X = 3 five times and X = 4 seven times.

P(1) = 116, P(2) = 316, P(3) = 516, P(4) = 716.

E(X) = 1 + 6 + 15 + 2816 = 5016 = 258 = 3.125.

Answer: E(X) = 258 = 3.125

Scoring: 1 point for the distribution, 1 point for the expected value.