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

Answer 2

The expected value of the prize won by a prospective customer receiving a flier is $7.4124. The standard deviation is $277.1.

What is a variance?

Variance is the value of the squared variation of the random variable from its mean value, in probability and statistics.

What is a standard deviation?

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

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.

A.  The probability will be

$\rm P(X=x_i) = p_i\\\\P[X = winning\ the\ car\ ( 32000)] = \dfrac{1}{13320} = 0.000075\\\\\\P[X = \rm winning\ the\ gas\ card\ ( 175)] = \dfrac{1}{13320} = 0.000075\\\\\\P[X = \rm winning\ the\ shopping\ car\ ( 5)] = \dfrac{13318}{13320} = 0.99985$

The expected value is given as

$\rm E(X) = \Sigma i_ip_i\\\\E(X) = (32000*0.000075) + (175*0.000075) + (5*0.99985)\\\\E(X) = \ 7.4124$

B.  The standard deviation will be

$\sigma = \sqrt{Variance}$

And Variance will be

$\rm Var(X) = \Sigma x^2 p- \mu ^2$

Where μ = E(X) = 7.4124

$\rm Var(X) = (32000^2*0.000075) + (175^2*0.000075) + (5^2 * 0.99985) - 7.4124^2 \\\\Var (X) = 76772.349$

Then the standard deviation will be

$\sigma = \sqrt{76772.349} = \ 277.1$

More about the standard deviation and the variance link is given below.

https://brainly.com/question/10687815