What are the roles of measure of dispersion in descriptive statistics? Following table gives the frequency distribution of thickness of computer chips (in nanometer) manufactured by two companies.

Thickness of computer chips 5 10 15 20 25 30
Number of chips Company A 10 15 24 20 18 13
company B 12 18 20 22 24 4

This answer is restricted. Please login to view the answer of this question.

Login Now

Measure of dispression is a descriotive statistical measure used to measure the variation or spreed or scatter less in the data set.

It gives additional information that enables to judge the reliability of measure of central tendency.

The measure of dispersion is important because it determines the margin of error. You’ll have when measuring the central tendency.

Solution Part:

For Company A:

x f fx fx2
5 10 50 250
10 15 150 1500
15 24 360 5400
20 20 400 10000
25 18 450 11250
30 13 390 11700
N = 100 Σfx = 1800 Σfx2 = 40100

Now,

\(\bar{x} = \frac{\sum fx}{N}\) = \(\frac{1800}{100}\) = 18

Standard deviation(σ) = \(\sqrt{\frac{\sum fx^2}{N} – \left ( \frac{\sum fx}{N} \right )^2}\)

= \(\sqrt{\frac{40100 }{100} – \left ( \frac{1800}{100} \right )^2}\)

= 8.77

And, Co-varience (c.v) = \(\frac{σ}{\bar{x}} \times 100%\)

= \(\frac{8.77}{18} \times 100%\)

= 48.72%

For Company b:

x f fx fx2
5 12 60 300
10 18 180 1800
15 20 300 4500
20 22 440 8800
25 24 600 15000
30 4 120 3600
N = 100 Σfx = 1700 Σfx2 = 34000

Now,

\(\bar{x} = \frac{\sum fx}{N}\) = \(\frac{1700}{100}\) = 17

Standard deviation(σ) = \(\sqrt{\frac{\sum fx^2}{N} – \left ( \frac{\sum fx}{N} \right )^2}\)

= \(\sqrt{\frac{34000}{100} – \left ( \frac{1700}{100} \right )^2}\)

= 7.14

And, Co-varience (c.v) = \(\frac{σ}{\bar{x}} \times 100%\)

= \(\frac{7.14}{17} \times 100%\)

= 42%

Since, the standard deviation and covarience of company B is less than that of company A. so, company B is considered more consistent in terms of thickness of computer chip.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .