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Magazine Chapter 1: Q21E (page 27)
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.
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