Question

The normal monthly precipitation (in inches) for September is listed for 20 different U.S. Cities.

3.5       1.6       2.4       3.7       4.1       3.9       1.0       3.6       4.2       3.4       3.7       2.2       1.5       4.2 3.4       2.7       0.4       3.7       2.0       3.6

Find

Mean of the data

Median of the data

Range of the data

Interquartile range of the data.   

Answer 1

We have been given the normal monthly precipitation of 20 different US cities.

We must find the mean, median, range and the interquartile range.

Mean:

The mean is the sum of all the observations, divided by the number of observations.

Sum= 3.5+1.6+2.4+3.7+4.1+3.9+1.0+3.6+4.2+3.4+3.7+2.2+1.5+4.2+3.4+2.7+0.4+3.7+2.0+3.6

= 58.8

Number of observations= 20

Thus, mean= 58.8 / 20

= 2.94

Median:

To find the median, we must first arrange the data in ascending order. The rearranged data is as follows:

0.4, 1.0, 1.5, 1.6, 2.0, 2.2, 2.4, 2.7, 3.4, 3.4, 3.5, 3.6, 3.6, 3.7, 3.7, 3.7, 3.9, 4.1, 4.2, 4.2

To find the median, we must select the 10th and the 11th observation (of the rearranged data) and divide it by 2.

Thus, median= (3.4+3.5) / 2

= 3.45

Range:

Range is the difference between the maximum value and the minimum value in the data.

Thus, range= 4.2-0.4

= 3.8

Interquartile range:

To find the interquartile range, we must find the first quartile(Q1) and the third quartile(Q3). The difference between Q3 and Q1 is the interquartile range.

Q1 is 5th observation+ the 6th observation, whole divided by 2.
Q3 is 15th observation+ the 16th observation, whole divided by 2.

Thus, Q1= (2.0+2.2) / 2
= 2.1

Q3= (3.7+3.7) / 2
= 3.7
Thus, interquartile range= 3.7-2.1
= 1.6