Chapter 1, Problem 1.48SE

a.

To determine

To find: The relative frequency histogram for the data.

Answer to Problem 1.48SE

The relative frequency histogram for the data is shown in Figure-1.

Explanation of Solution

Given information: The data is shown below.

8.9	12.3	10.7	8.4	7.8	11.5	8.2	9.0	8.8
7.1	11.8	10.3	7.7	9.0	10.5	9.1	8.7	8.7
9.1	9.0	10.5	11.2	7.7	8.8	12.2	7.9	8.8
8.7	7.1	8.7	7.6	5.1	35.1	10.5	10.4	11.0
10.2	8.6	10.7	9.6	8.3	8.0	9.5	7.7	9.4
8.7	7.8	10.2	6.9	9.2	10.2	6.2	9.6	12.2

Calculation:

The histogram is shown below.

  

Figure-1

Thus, the histogram for the data is shown in Figure-1.

b.

To determine

To explain:Whether the observation describes the geography of the city.

Explanation of Solution

Given information: The data is shown below.

8.9	12.3	10.7	8.4	7.8	11.5	8.2	9.0	8.8
7.1	11.8	10.3	7.7	9.0	10.5	9.1	8.7	8.7
9.1	9.0	10.5	11.2	7.7	8.8	12.2	7.9	8.8
8.7	7.1	8.7	7.6	5.1	35.1	10.5	10.4	11.0
10.2	8.6	10.7	9.6	8.3	8.0	9.5	7.7	9.4
8.7	7.8	10.2	6.9	9.2	10.2	6.2	9.6	12.2

Since, the highest peak in the Mount Washington is at 6288 ft which is famous for the erratic weather but the recorded value is an outlier with respect to the geographical location of the city.

c.

To determine

To explain:Whether the wind is unusual.

Explanation of Solution

Given information: The data is shown below.

8.9	12.3	10.7	8.4	7.8	11.5	8.2	9.0	8.8
7.1	11.8	10.3	7.7	9.0	10.5	9.1	8.7	8.7
9.1	9.0	10.5	11.2	7.7	8.8	12.2	7.9	8.8
8.7	7.1	8.7	7.6	5.1	35.1	10.5	10.4	11.0
10.2	8.6	10.7	9.6	8.3	8.0	9.5	7.7	9.4
8.7	7.8	10.2	6.9	9.2	10.2	6.2	9.6	12.2

Since, the relative frequency is $0.189$ and the value is not a extreme value.

Thus, the observation is usually windy.