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An average is a value that is typical of a larger data set, there are three common averages; the mean, median and mode. Each  of these averages represents a central value around which the given data are distributed, thus they are often called measures of central tendency.

Mean:

The mean, or arithmetic mean is the most common kind of average. If there are “n” numbers in a set then

Mean = \( \frac{sum\;of\;the\;numbers\;in\;a\; set}{n}\)

i.e. Mean = X =∑ \( \scriptsize \frac{x}{n} \)

Median:

The median is gotten by arranging data in order of size, then the middle term is called the median. If there is an even number of terms the median is the arithmetic mean of the two middle terms.

Mode:

The mode of a set of numbers is the number which appears most often. i.e. the number with the greatest frequency.

Comparison of Mean, Median and Mode

The mean is the most widely used average. However, in a distribution which contains extreme values, the mean may not be typical of the data. Thus, in distribution where there are extreme values, the median is the most commonly used average.

The mode is commonly used in the mass-production industries where production is geared towards the most popular kinds of consumer items, such as clothes, cars and newspapers.

Example 1

The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows:

Screen Shot 2020 10 14 at 10.27.03 PM

(a) Construct a frequency distribution table using class intervals 45 – 54, 55-64, etc.

(b) Using an assumed mean of 77.5, find the mean of the distribution.

(c) Find the mode of the distribution by means of formula.

(d) Draw the histogram of the distribution

(e) Use your histogram to estimate the mode of the distribution. (King’s College, Lagos)

Solution

BirthBoundariesFrequency (f)Middle class (x)Deviation (d)f(x)
45 – 5444.5 – 54.5149.5-2849.5
55 – 6454.5 -64.5259.5-18119
65 – 7464.5 – 74.51569.5-81,042.5
75 – 8474.5 – 84.52179.521,669.5
85 – 9484.5 – 94.5789.512626.5
95 – 10494.5 – 104.5399.522298.5
105 – 114104.5 – 114.51109.532109.5
∑f=50∑d=143,915

(b) Mean = \( \scriptsize \bar{X}\)= Assumed mean + \( \frac {∑d}{∑f}\)

= 77.5 + \( \frac{14}{50} \)

= 77.5 + 0.28

= 77.78

Therefore, mean \( \scriptsize \bar{X}\) = 78

(Recall ∑f =50 i.e. this is a discrete distribution, thus the mean must be discrete i.e. 78).

(c) Mode = \( \scriptsize L + \frac {C \left (fm \;- \;fb \right)}{\left (fm \;- \; fb \right) + \left (fm \;- \; fa \right)} \)

Where L = lower class boundary of the modal class

C = width of the class interval

fm = frequency of modal class

fb =frequency of class before the modal class

fa = frequency of class after the modal class

The modal class is 75 – 84

i.e. L = 75 – 0.5 = 74.5

C = 84.5 – 74.5 = 10

fm = 21

fb = 15

Therefore, Mode = 74.5 + \( \frac {10 \left (21 \;- \; 15 \right)}{\left (21 \;- \; 15 \right) + \left (21 \;- \; 7 \right)} \)

=74.5 + \( \frac{10(6)}{6 + 14} \)

= 74.5 + \( \frac{60}{20} \)

= 74.5 + 3

= 77.5

= 78 (for discrete distribution)

i.e. Mode = 78 births.

= 77.5

=78 (for discrete distribution)

i.e. Mode = 78 births.

histogram e.g

From the histogram the mode is estimated to be 77.5

i.e. Mode = 78 births (discrete distribution).

Example 2

The densities \( \frac{Kg}{m^3} \) of 80 items give the following distribution

Densities \( \frac{Kg}{m^3} \)Frequency
11 – 205
21 – 307
31 – 4012
41 – 5014
51 – 608
61 – 704

(i) Draw the histogram of the distribution

(ii) Find the modal class of the distribution

(iii) Use the histogram to estimate the mode of the distribution

(iv) Calculate the mode of the distribution using the formula for mode.

Solution:

(i) Densities of Items

Screen Shot 2020 10 14 at 10.41.58 PM

(ii) From the frequency table, the modal class is 41 – 50.

(iii) From the histogram the mode of the distribution is estimated from the intersection from the diagonals from the adjacent bars to the modal class – this is estimated to be 43Kgm3

(iv) Mode = Mode = \( \scriptsize L + \frac {C \left (fm \;- \;fb \right)}{\left (fm \;- \; fb \right) + \left (fm \;- \; fa \right)} \)

Where L = lower class boundary of the modal class = 40.5

C = width of the class interval = 50.5 – 40.5 = 10

fm = frequency of modal class = 14

fb = frequency of class before the modal class = 12

fa = frequency of class after the modal class=

Therefore, Mode = 40.5 + \( \frac {10 \left (14 \;- \; 12 \right)}{\left (14 \;- \; 12 \right) + \left (14 \;- \; 8 \right)} \)

Mode = 40.5 + \( \frac{20}{8} \)

Mode = 40.5 + 2.5

= 43\( \frac{Kg}{m^3} \)

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