This problem involves measures of dispersion rather than just measures of centra

Question

This problem involves measures of dispersion rather than just measures of central tendency. The newspaper The Los Angeles Times regularly reports the index of air quality for several regions of Southern California. A sample of the air quality drawn on different days for the city of Pomona gave the following values: 28, 42, 58, 48, 45, 55, 60, 49, and 50.

a) Calculate the range for the entire sample as well as the interquartile range.
b) Calculate the variance and the standard deviation of the sample.
c) Calculate the sum of the deviations from the mean value raised to only the first power. Perhaps you now understand why the variance is a quadratic form, meaning that it involves the sum of squared quantities.
d) A sample of these index values for air quality in the nearby city of Anaheim gives a mean value of 48.5, a variance of 136, and a standard-deviation of 11.66. Based on these descriptive statistics, what comparisons would you draw in regards to the air quality in these two cities?

Explanation / Answer

after arranging the data in ascending order we generated following information

(a) range=maximum-minimum=60-28=32

Interquartile range=Q3-Q1=55-45=10

first quartile Q1=45 and third quartile Q3=55

(b) variance=sum(x-mean)2/n=742/9=82.44

standard deviation=sqrt(variance)=sqrt(82.44)=9.08

(c) sum of deviation from mean=0, it is calculated on above table

(d) Anaheim and Pomona have almost same mean value but variability or heteroginity in Anaheim is more as its variance is more.

s.n. x x-mean (x-mean)2 1 28 -20.33 413.44 2 42 -6.33 40.11 3 45 -3.33 11.11 4 48 -0.33 0.11 5 49 0.67 0.44 6 50 1.67 2.78 7 55 6.67 44.44 8 58 9.67 93.44 9 60 11.67 136.11 mean= 48.33 0.00 82.44 sum= 435 0 742 n= 9 9 9

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