Chapter 5.3, Problem 18E

The American Housing Survey reported the following data on the number of bedrooms in owner-occupied and renter-occupied houses in central cities (U.S. Census Bureau website, March 31, 2003).

	Number of Houses (1000s)
Bedrooms	Renter-Occupied	Owner-Occupied
0	547	23
1	5012	541
2	6100	3832
3	2644	8690
4 or more	557	3783

1. a. Define a random variable x = number of bedrooms in renter-occupied houses and develop a probability distribution for the random variable. (Let x = 4 represent 4 or more bedrooms.)
2. b. Compute the expected value and variance for the number of bedrooms in renter-occupied houses.
3. c. Define a random variable y = number of bedrooms in owner-occupied houses and develop a probability distribution for the random variable. (Let y = 4 represent 4 or more bedrooms.)
4. d. Compute the expected value and variance for the number of bedrooms in owner-occupied houses.
5. e. What observations can you make from a comparison of the number of bedrooms in renter-occupied versus owner-occupied homes?

To determine

Construct a discrete probability distribution for the random variable x.

Answer to Problem 18E

The probability distribution for the random variable x is given by,

x	f(x)
0	0.04
1	0.34
2	0.41
3	0.18
4	0.04

Explanation of Solution

Calculation:

The data represents the number of bedrooms in owner-occupied and renter-occupied houses. The random variable x is the number of bedrooms inrenter-occupied houses.

The corresponding probabilities of the random variable x are obtained by dividing the number of houses that are renter occupied (f) with totalnumber of houses (N) that are renter occupied.

That is, $f\left(x\right)=\frac{f}{N}$

For example, $f\left(0\right)=0.04\left(=\frac{547}{14,860}\right)$

The probability distribution for the random variable x can be obtained as follows:

x	f	f(x)
0	547	0.04
1	5,012	0.34
2	6,100	0.41
3	2,644	0.18
4	557	0.04
Total	14,860	1.00

Thus, a discrete probability distribution for the random variable x is obtained.

To determine

Compute the expected value and variance for the random variable x.

Answer to Problem 18E

The expected value for the random variable x is1.84.

The variance of the random variable x is 0.79.

Explanation of Solution

Calculation:

The formula for the expected value of a discrete random variable is,

$\begin{array}{c}E\left(x\right)=\mu \\ ={\displaystyle \sum x\cdot f\left(x\right)}\end{array}$

The formula for the variance of the discrete random variable is,

${\sigma }^2={\displaystyle \sum \left[{\left(x-\mu \right)}^2.f\left(x\right)\right]}$

The expected value and variance for the random variable x is obtained using the following table:

x	f(x)	$x\cdot f\left(x\right)$	$\left(x-\mu \right)$	${\left(x-\mu \right)}^2$	${\left(x-\mu \right)}^2.f\left(x\right)$
0	0.04	0.00	–1.84	3.39	0.12
1	0.34	0.34	–0.84	0.71	0.24
2	0.41	0.82	0.16	0.02	0.01
3	0.18	0.53	1.16	1.34	0.24
4	0.04	0.15	2.16	4.66	0.17
Total	1.0000	1.84			0.79

Thus, the expected value for the random variable x is 1.84 and the variance of the random variable x is 0.79.

To determine

Construct a discrete probability distribution for the random variable y.

Answer to Problem 18E

The probability distribution for the random variable y is given by,

y	f(y)
0	0.00
1	0.03
2	0.23
3	0.52
4	0.22

Explanation of Solution

Calculation:

The given information is that the random variable y represents the number of bedrooms in owner-occupied houses.

The corresponding probabilities of the random variable y are obtained by dividing the number of houses that are owner occupied (f) with total number of houses(N) that are owner occupied.

That is, $f\left(y\right)=\frac{f}{N}$

For example, $f\left(0\right)=0.00\left(=\frac{23}{16,869}\right)$

The probability distribution for the random variable y can be obtained as follows:

y	f	f(y)
0	23	0.00
1	541	0.03
2	3,832	0.23
3	8,690	0.52
4	3,783	0.22
Total	16,869	1.00

Thus, a discrete probability distribution for the random variable y is obtained.

To determine

Compute the expected value and variance for the random variable y.

Answer to Problem 18E

The expected value for the random variable y is 2.93.

The variance of the random variable y is 0.59.

Explanation of Solution

Calculation:

The formula for the expected value of a discrete random variable is,

$\begin{array}{c}E\left(y\right)=\mu \\ ={\displaystyle \sum y\cdot f\left(y\right)}\end{array}$

The formula for the variance of the discrete random variable is,

${\sigma }^2={\displaystyle \sum \left[{\left(y-\mu \right)}^2.f\left(y\right)\right]}$

The expected value and variance for the random variable y is obtained using the following table:

y	f(y)	$y\cdot f\left(y\right)$	$\left(y-\mu \right)$	${\left(y-\mu \right)}^2$	${\left(y-\mu \right)}^2.f\left(y\right)$
0	0.00	0.00	–2.93	8.58	0.01
1	0.03	0.03	–1.93	3.72	0.12
2	0.23	0.45	–0.93	0.86	0.20
3	0.52	1.55	0.07	0.01	0.00
4	0.22	0.90	1.07	1.15	0.26
Total	1.00	2.93			0.59

Thus, the expected value for the random variable y is 2.93 and the variance of the random variable y is 0.59.

To determine

Explain the observations that can be made from a comparison of the number of bedrooms in renter occupied versus owner occupied homes.

Explanation of Solution

The expected number of bedrooms in renter-occupied houses is 1.84 and the expected number of bedrooms in owner-occupied houses is 2.93. Thus, the expected value is greater for the owner-occupied houses than the renter-occupied houses. The variability is less for the owner-occupied houses comparing to the renter-occupied houses.