Question

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three different​ prizes: an automobile valued at ​$32,000, a ​$175 gas​ card, or a ​$5 shopping card. To claim his or her​ prize, a prospective customer needed to present the flier at the​ dealership's showroom. The fine print on the back of the flier listed the probabilities of winning The chance of winning the car was 1 out of 13,320, the chance of winning the gas card was 1 out of 13,320, and the chance of winning the shopping card was 13,318 out of 13,320. Complete parts (a) and (b).
(a). Using the probabilities listed on the flier, what is the expected value of the prize won by a prospective customer receiving a flier?
(b). Using the probabilities listed on the flier, what is the standard deviation of the value of the prize won by a prospective customer receiving a flier?

Answer 1

Answer:

a) E(X) = $7.4124

b) Standard deviation = $277.1

Step-by-step explanation:

P(X=xᵢ) = pᵢ

P[X=winning the car($32000)]= 1/13320 = 0.000075

P[X=winning the gas card($175)] = 1/13320 = 0.000075

P[X=winning the shopping car($5)] = 13318/13320 = 0.99985

Expected value is given by

E(X) = Σ xᵢpᵢ

E(X) = (32000 × 0.000075) + (175 × 0.000075) + (5 × 0.99985) = $7.412375 = $7.4124

b) Standard deviation = √(variance)

But Variance = Var(X) = Σx²p − μ²

where μ = E(X) = 7.4124

Σx²p = (32000² × 0.000075) + (175² × 0.000075) + (5² × 0.99985) = 76,827.29

Var(X) = 76,827.29 - 7.4124² = 76,772.349

Standard deviation = √(76,772.349) = $277.1