A Gallup Daily Tracking Survey found that the mean daily discretionary spending

Question

A Gallup Daily Tracking Survey found that the mean daily discretionary spending by Americans earning over $90,000 per year was $136 per day (USA Today, July 30, 2012). The discretionary spending excluded home purchases, vehicle purchases, and regular monthly bills. Let  and assume that a uniform probability density function applies with  for .

Find the values of a and b for the probability density function.

What is the probability that consumers in this group have daily discretionary spending between $100 and $200?

What is the probability that consumers in this group have daily discretionary spending of $150 or more?

What is the probability that consumers in this group have daily discretionary spending of $80 or less?

Explanation / Answer

1)
f(x) = 1/(b-a), so b-a = 1/0.00625 = 160.

The mean is 136 = (b+a)/2, so b+a = 272.

Adding these two equations, we get 2b=432, so
b=216 and
a=216-160 = 56

so, (a, b) = (56, 216)

2)

P(c<X<d) = dc/ba

P(100<X<200) = 200-100/216-56 = 0.625

3)

P(150<=X<=216) = 216-150/216-56 = 0.413

4)

P(56<=X<=80) = 80-56/216-56 = 0.15

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