Question

According to the U.S. Department of Labor, the average American household spends $639 on household supplies per year. Suppose annual expenditures on household supplies per household are uniformly distributed between the values of $251 and $1,027.
(a) What is the standard deviation of this distribution?
(b) What is the height of this distribution?
(c) What proportion of households spend more than $870 per year on household supplies?
(d) What proportion of households spend more than $1,290 per year on household supplies?
(e) What proportion of households spend between $380 and $490 on household supplies?

Answer 1

Answer:

a) $Var(X) = \frac{(b-a)^2}{12} = \frac{(1027 -251)^2}{12}= 50181.333$

And the deviation would be:

$Sd(X) = \sqrt{50181.333}= 224.012$

b) $h = \frac{1}{b-a} =\frac{1}{1027-251}=0.00129$

c) $P(X>870) = 1-P(X<870) = 1- \frac{870-251}{1027-251}= 0.2023$

d) $P(X>1290) = 1-P(X<1290) = 1- \frac{1027-251}{1027-251}= 0$

e) $P(380<X<490) = P(X<490)-P(X<380) = \frac{490-251}{1027-251}- \frac{380-251}{1027-251}= 0.3080-0.1662= 0.1418$

Step-by-step explanation:

For this case we define the random variable X as "American household spend" and we know that the distribution for X is given by:

$X \sim Unif (a=251, b =1027)$

Part a

We can calculate the variance first with this formula:

$Var(X) = \frac{(b-a)^2}{12} = \frac{(1027 -251)^2}{12}= 50181.333$

And the deviation would be:

$Sd(X) = \sqrt{50181.333}= 224.012$

Part b

For this case the height represent the individual probability for any value in the interval and is given by:

$h = \frac{1}{b-a} =\frac{1}{1027-251}=0.00129$

Part c

For this case we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a} , a \leq X \leq b$

We want this probability:

$P(X>870) = 1-P(X<870) = 1- \frac{870-251}{1027-251}= 0.2023$

Part d

For this case we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a} , a \leq X \leq b$

We want this probability:

$P(X>1290) = 1-P(X<1290) = 1- \frac{1027-251}{1027-251}= 0$

Part e

We want this probability:

$P(380<X<490) = P(X<490)-P(X<380) = \frac{490-251}{1027-251}- \frac{380-251}{1027-251}= 0.3080-0.1662= 0.1418$