Question

The article cited in Exercise 20 also gave the following values of the variables y=number of culs-de-sac and z=number of intersections:

y	1	0	1	0	0	2	0	1	1	1	2	1	0	0	1	1	0	1	1
z	1	8	6	1	1	5	3	0	0	4	4	0	0	1	2	1	4	0	4

y	1	1	0	0	0	1	1	2	0	1	2	2	1	1	0	2	1	1	0
z	0	3	0	1	1	0	1	3	2	4	6	6	0	1	1	8	3	3	5

y	1	5	0	3	0	1	1	0	0
z	0	5	2	3	1	0	0	0	3

a. Construct a histogram for the ydata. What proportion of these subdivisions had no culs-de-sac? At least one cul-de-sac?

Short answer

a.

The histogram is represented as,

Theproportion ofsubdivisions that had no culs-de-sac is 0.362.

The proportion of subdivisions that had at least oneculs-de-sac is 0.638.

Step-by-step solution

Given information

It is given that y represents the number of culs-de-sac and z represents the number of interactions.

Construct a histogram.

a.

The total number of culs-de-sac is 47.

The relative frequency is computed as,

\({\rm{relative frequency }} = \frac{{frequency}}{{Total\;number\;of\;observations}}\)

The table representing the relative frequency is computed as,

y	Frequency	Relative frequency
0	17	0.362
1	22	0.468
2	6	0.128
3	1	0.021
4	0	0.000
5	1	0.021

Steps to construct a histogram are,

1) Determine the frequency or the relative frequency.

2) Mark the class boundaries on the horizontal axis.

3) Draw a rectangle on the horizontal axis corresponding to the frequency or relative frequency.

The histogram is represented as,

Compute the proportion

The proportionof subdivisions that had no culs-de-sac is computed as,

\(P\left( {y = 0} \right) = 0.362\)

Therefore, theproportion ofsubdivisions that had no culs-de-sac is 0.362.

The proportionof subdivisions that had at least one culs-de-sac is computed as,

\(\begin{aligned}P\left( {y \ge 1} \right) &= 1 - P\left( {y = 0} \right)\\ &= 1 - 0.362\\ &= 0.638\end{aligned}\)

Therefore, the proportion of subdivisions that had at least one culs-de-sac is 0.638.