The article \"Determination of Most Representative Subdivision\"† gave data on v

Question

The article "Determination of Most Representative Subdivision"† gave data on various characteristics of subdivisions that could be used in deciding whether to provide electrical power using overhead lines or underground lines. Here are the 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 1 1 0 0 0   z   1 8 6 1 1 5 3 0 0 4 4 0 0 1 2 1 4 0 4 0 3 0 1 1   y   1 1 2 0 1 2 2 1 1 0 2 1 1 0 1 5 0 3 0 1 1 0 0   z   0 1 3 2 4 6 6 0 1 1 8 3 3 5 0 5 2 3 1 0 0 0 3 Number of Number of 0 4 5 culs-de-sac 0 4 5 culs-de-sac What proportion of these subdivisions had no culs-de-sac? At least one cul-de-sac? (Round your answers to three decimal places.) no culs-de-sac at least one cul-de-sac (b) Construct a histogram for the z data. Frequency Frequency 14 15 12 10 10 Number of Number of 0 1 2 3 4 5 6 7 8 intersections 0 1 2 3 4 5 6 7 8 intersections Frequency Frequency 14 12 12 10 10 Number of Number of 0 1 2 3 4 5 6 7 8 intersections 0 1 2 3 4 5 6 7 8 intersections hat proportion of the subdivisions had at most five intersections? Fewer than five intersections? (Round your answers to three decimal places.) at most fiver intersections fewer than five intersections 0.8994

Explanation / Answer

(A)

Out of 47 observations, 17 has no clue-de-sac so the required proprotion is

P(no clus-de-sac) = 17/ 47 = 0.362

and by the complement rule

P(at least one clue-de-sac) = 1- 0.362 = 0.638

(B)

Out of 47 subdivisions, 42 has at most 5 intersections so

P(at most five intersections) = 42/47 = 0.894

and Out of 47 subdivisions, 39 has fewer intersections so

P(fewer than five intersections) = 39/47 = 0.830

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