Mathematics

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Calculation of variances and standard deviation

For discrete variables, variance is calculated using the formula: Where X are the individual scores, is the mean of the data set and n is the number of values in the data set. The standard deviation is obtained is obtained by getting the square root of the variances, hence; When a frequency table is used, the variance can also be given by the formula:

           or              

Where ƒ is the frequency of each individual value and ∑ƒ is the sum of the individual frequencies which is the same as n. For grouped or continuous data the same formula is used but the X value is taken to be mid-point of each group. For example, if the groups are 10-14, 15-19, 20-24 and so on, then the mid-points will be 12, 17 and 22 respectively. These become the class representatives and thus assume the value of X in the calculation. Example The frequency distribution table below shows the number of hours that a group of 220 students spent in private study within a given month. Number of hours per week spent watching television

Hours Number of students

10–14 2

15–19 12

20–24 23

25–29 60

30–34 77

35–39 38

40–44 8

 (a)	Find:	(i)	The mean of the distribution

 (ii)The standard deviation of the distribution

 (b)	Assuming that the distribution is normal, calculate the percentage of students who studied for 35.85 hours or more. 
       Solutions

In order to work out these values it is appropriate to draw up a tabular format on which to enter the necessary information.

Hours Midpoint(x) Frequency(f) xf (x - ) (x - )2 f(x - )2 10–14 12 2 24 -17.82 317.6 635.2 15–19 17 12 204 -12.82 164.4 1,972.8 20–24 22 23 506 -7.82 61.2 1,407.6 25–29 27 60 1,620 -2.82 8.0 480.0 30–34 32 77 2,464 2.18 4.8 369.6 35–39 37 38 1,406 7.18 51.6 1,960.8 40–44 42 8 336 12.18 148.4 1,187.2     220 6,560     8,013.2

(a)(i) Mean

= 6560 220 = 29.82

(ii) Standard deviation

S =

  	  =    6.03

(b)Percentage who studied for 35.85 hours or more: On examination, we see that 35.85 is the same as the mean (29.82) plus one standard deviation (6.03). Thus the percentage we want is that on the normal curve beyond +1σ from the mean. From the normal curve we get: 13.6% + 2.15% + 0.15% = 15.9% Thus we can say that 15.9% of the students in that class studied for 35.85 hours or more during the given month. Note that in this case, normality of the distribution is only an assumption to enable us do statistical analysis. Such assumptions are done often since in real life it is rare to come across perfectly normal distributions. With today’s advancement in computing technology, the researcher does not have to go through the rigours of calculating such statistical measures. Ordinary calculators and computers can easily give the values when fed with the correct details.

• * Free High School Mathematics Textbook (Gr. 10-12)], Shuttleworth Foundation, South Africa.