Chapter 5, Problem 93CE

Four students divided the task of surveying the types of vehicles in parking lots of four different shopping malls. Each student examined 100 cars in each of four large suburban malls, resulting in the 5 × 4 contingency table shown below. (a) Calculate each probability (i–vi) and explain in words what it means. (b) Do you see evidence that vehicle type is not independent of mall location? Explain. (Data are from an independent project by MBA students Steve Bennett, Alicia Morais, Steve Olson, and Greg Corda.)

1. i. P(C)
2. ii. P(G
3. iii. P(V | S)
4. iv. P(C | J)
5. v. P(C and G)
6. vi. P(T and O)

To determine

Calculate each probability (i-vi) and explain it meaning.

Answer to Problem 93CE

i. The probability $P\left(C\right)$ is 0.4825 and the meaning is the probability of seeing a car in a shopping mall parking lot is 0.4825.

ii. The probability $P\left(G\right)$ is 0. 25 and the meaning is the probability of seeing a vehicle in the Greek Lakes shopping mall is 0.25.

iii. The probability $P\left(V|S\right)$ is 0.19 and the meaning is the probability of seeing a parked SUV at the Somerset mall is 0.19.

iv. The probability $P\left(C|J\right)$ is 0.64 and the meaning is the probability of seeing a parked car at the Jamestown mall is 0.64.

v. The probability $P\left(C\text{ and G}\right)$ is 0.09 and the meaning is the probability of parked vehicle is a car and is at the Great Lakes mall is 0.09.

vi. The probability $P\left(T\text{ and O}\right)$ is 0.015 and the meaning is the probability of parked vehicle is a truck and is at the Oakland mall is 0.015.

Explanation of Solution

Calculation:

The given table shows that number of vehicles of each type in four shopping malls.

The given contingency table is,

Vehicle Type	Somerset (S)	Oakland (O)	Great Lakes (G)	Jamestown (J)	Row Total
Car (C)	44	49	36	64	193
Minivan (M)	21	15	18	13	67
Full-size van (F)	2	3	3	2	10
SUV (V)	19	27	26	12	84
Truck (T)	14	6	17	9	46
Column Total	100	100	100	100	400

For (i)$P\left(C\right)$:

The formula for finding the probability $P\left(C\right)$ is,

$P\left(C\right)=\frac{\text{Frequency for the class }C}{\text{Total frequencies in the distribution}}$

Substitute 193 for ‘Frequency for the class C’ and 400 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(C\right)=\frac{193}{400}\\ =0.4825\end{array}$

Therefore, the probability $P\left(C\right)$ is 0.4825 and the meaning is the probability of seeing a car in a shopping mall parking lot is 0.4825.

For (ii) $P\left(G\right)$:

The formula for finding the probability $P\left(G\right)$ is,

$P\left(G\right)=\frac{\text{Frequency for the class }G}{\text{Total frequencies in the distribution}}$

Substitute 100 for ‘Frequency for the class G’ and 400 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(G\right)=\frac{100}{400}\\ =0.25\end{array}$

Therefore, the probability $P\left(G\right)$ is 0.25 and the meaning is the probability of seeing a vehicle in the Greek Lakes shopping mall is 0.25.

For (iii) $P\left(V|S\right)$:

The formula for finding the probability $P\left(V|S\right)$ is,

$P\left(V|S\right)=\frac{\text{Frequency for the class }V\text{ and }S}{\text{Frequency for the class }S}$

Substitute 19 for ‘Frequency for the class V and S’ and 100 for ‘Frequency for the class S’,

$\begin{array}{c}P\left(V|S\right)=\frac{19}{100}\\ =0.19\end{array}$

Therefore, the probability $P\left(V|S\right)$ is 0.19 and the meaning is the probability of seeing a parked SUV at the Somerset mall is 0.19.

For (iv) $P\left(C|J\right)$:

The formula for finding the probability $P\left(C|J\right)$ is,

$P\left(C|J\right)=\frac{\text{Frequency for the class }C\text{ and }J}{\text{Frequency for the class }J}$

Substitute 19 for ‘Frequency for the class C and J’ and 100 for ‘Frequency for the class J’,

$\begin{array}{c}P\left(C|J\right)=\frac{64}{100}\\ =0.64\end{array}$

Therefore, the probability $P\left(C|J\right)$ is 0.64 and the meaning is the probability of seeing a parked car at the Jamestown mall is 0.64.

For (v) $P\left(C\text{ and G}\right)$:

The formula for finding the probability $P\left(C\text{ and G}\right)$ is,

$P\left(C\text{ and G}\right)=\frac{\text{Frequency for the class }C\text{ and }G}{\text{Total frequencies in the distribution}}$

Substitute 36 for ‘Frequency for the class C and G’ and 400 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(C\text{ and G}\right)=\frac{36}{400}\\ =0.09\end{array}$

Therefore, the probability $P\left(C\text{ and G}\right)$ is 0.09 and the meaning is the probability of parked vehicle is a car and is at the Great Lakes mall is 0.09.

For (vi) $P\left(T\text{ and O}\right)$:

The formula for finding the probability $P\left(T\text{ and O}\right)$ is,

$P\left(T\text{ and O}\right)=\frac{\text{Frequency for the class }T\text{ and }0}{\text{Total frequencies in the distribution}}$

Substitute 6 for ‘Frequency for the class T and O’ and 400 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(T\text{ and O}\right)=\frac{6}{400}\\ =0.015\end{array}$

Therefore, the probability $P\left(T\text{ and O}\right)$ is 0.015 and the meaning is the probability of parked vehicle is a truck and is at the Oakland mall is 0.015.

To determine

Check whether there is evidence that vehicle type is not independent of mall location and explain.

Answer to Problem 93CE

Yes, there is evidence that vehicle type is not independent because $P\left(T\text{ and }O\right)\ne P\left(T\right)P\left(O\right)$.

Explanation of Solution

Calculation:

Special law of multiplication:

If two events A and B are independent, then

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$.

Consider vehicle type truck and mall location as Oak land.

From part (a), $P\left(T\text{ and O}\right)=0.015$.

The formula for checking whether there is evidence that vehicle type is not independent of mall location or not is,

$\begin{array}{c}P\left(T\cap O\right)=P\left(T\right)P\left(O\right)\\ 0.015=\left(\frac{\text{Frequency for the class }T}{\text{Total frequencies in the distribution}}\right)\left(\frac{\text{Frequency for the class }O}{\text{Total frequencies in the distribution}}\right)\\ =\frac{46}{400}\times \frac{100}{400}\\ =\frac{4,600}{160,000}\\ 0.015\ne 0.0287\end{array}$

Here, it is observed that $P\left(T\text{ and }O\right)\ne P\left(T\right)P\left(O\right)$. Therefore, the there is evidence that vehicle type is not independent.