Question

1. an arcade game involves a player attempting to spin a wheel so that the pointer lands in the prize space when it comes to rest. there are five spaces on the wheel, including the prize space, and each has an arc length of 5 inches. the wheel has been programmed so that a spin is equally likely to land in any of the five spaces. let d = the distance past the center of the price space where the pointer stops after a spin. negative values indicate that the pointer stopped before the prize space. a uniform probability distribution modeling the continuous random variable d is shown, with a constant height of h. a. find the value of h. b. find ( p(d geq 5) ). c. find and interpret ( p(-2.5 leq d leq 2.5) ).

Explanation:

Step1: Calculate total interval width

The range of $D$ is from $-12.5$ to $12.5$.
Width = $12.5 - (-12.5) = 25$

Step2: Solve for $h$ (uniform distribution area = 1)

For a uniform distribution, area = width $\times h = 1$.
$25h = 1 \implies h = \frac{1}{25} = 0.04$
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Step3: Define interval for $P(D\geq5)$

The interval for $D\geq5$ is from $5$ to $12.5$.
Width = $12.5 - 5 = 7.5$

Step4: Calculate probability for part b

Probability = width $\times h$
$P(D\geq5) = 7.5 \times 0.04 = 0.3$
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Step5: Define interval for $P(-2.5\leq D\leq2.5)$

Width = $2.5 - (-2.5) = 5$

Step6: Calculate probability for part c

Probability = width $\times h$
$P(-2.5\leq D\leq2.5) = 5 \times 0.04 = 0.2$
Interpretation: There is a 20% chance the pointer stops within 2.5 inches of the center of the prize space.

Answer:

a. $\boldsymbol{0.04}$
b. $\boldsymbol{0.3}$
c. $\boldsymbol{0.2}$; There is a 20% probability that the pointer stops within 2.5 inches of the center of the prize space.