91影视

The data on the lengths of bus routes for any particular transit system will typically vary from one route to another is provided.

a.

The table is given as,

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,

b.

The relative frequency is computed as,

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

The table representing relative frequencies is given as,

The proportion of the route lengths that are less than 20 is computed as,

\(\begin{array}{c}P\left( {x < 20} \right) &=& P\left( {6 - < 8} \right) + P\left( {8 - < 10} \right) + P\left( {10 - < 12} \right) + ... + P\left( {18 - < 20} \right)\\ &=& 0.0153 + 0.0588 + 0.0767 + ... + 0.1074\\ &=& 0.5524\end{array}\)

Therefore, the proportion of the route lengths that are less than 20 is 0.5524.

The proportion of the route lengths that are of at least30 is computed as,

\(\begin{array}{c}P\left( {x \ge 30} \right) &=& P\left( {30 - < 35} \right) + P\left( {35 - < 40} \right) + P\left( {40 - < 45} \right)\\ &=& 0.0691 + 0.0281 + 0.0051\\ &=& 0.1023\end{array}\)

Therefore, the proportion of the route lengths that are of at least 30 is 0.1023.

c.

The 90^th percentile is computed as,

\({{\bf{P}}_{{\bf{90}}}}{\bf{ = l + }}\frac{{\bf{h}}}{{\bf{f}}}\left( {\frac{{{\bf{90N}}}}{{{\bf{100}}}}{\bf{ - C}}} \right)\)

Where,

l represent the lower limit of the percentile class.

h represents the class width.

C represents the prior cumulative frequency of the percentile class.

The relative frequency is computed as,

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

The cumulative frequency is computed by adding the prior class frequencies.

The table representing calculations is given as,

The size of the sample is\(N = 391\).

The percentile class is computed as,

\(\frac{{90N}}{{100}} = 351.9\)

It can be observed that the cumulative frequency greater than 351.9 is 378. Therefore, the percentile class is 30-35 with a corresponding frequency of 27.

The 90^th percentile is computed as,

\(\begin{array}{c}{P_{90}} &=& l + \frac{h}{f}\left( {\frac{{90N}}{{100}} - C} \right)\\ &=& 30 + \frac{5}{{27}}\left( {351.9 - 351} \right)\\ &=& 30 + 0.167\\ &=& 30.167\end{array}\)

Therefore, the 90^th percentile is 30.167.

Referring to the table represented in part c,

d.

The median is computed as,

\(Median = l + \frac{h}{f}\left( {\frac{N}{2} - C} \right)\)

Where,

I represent the lower limit of the percentile class.

h represents the class width.

C represents the prior cumulative frequency of the percentile class.

The median class is computed as,

\(\frac{N}{2} = 195.5\)

It can be observed that the cumulative frequency greater than 195.5 is 216. Therefore, the median class is 18-20 with a corresponding frequency of 42.

The median is given as,

\(\begin{array}{c}Median &=& l + \frac{h}{f}\left( {\frac{N}{2} - C} \right)\\ &=& 18 + \frac{5}{{42}}\left( {195.5 - 174} \right)\\ &=& 19.0234\end{array}\)

Therefore, the median route length is 19.024.