Calculate i. Mean ii. Median iii. Modal Class iv. Standard Deviation

Question

The mark students of grade 10 students got in their biology examination is given below

Marks (%)	Frequency
40-49	3
50-59	5
60-69	7
70-79	6
80-89	12
90-100	7

Calculate

i. Mean

ii. Median

iii. Modal Class

iv. Standard Deviation

Answer 1

Ali

We can only approximately calculate the mean, median, and standard deviation, as we do not have the exact grades, only the grade classes.

Let's start with (iii). The modal class is the class with the highest number of grades, so clearly that is 80-89.

To calculate the others we need to assume something about our data. Let's assume that all the values in a class are in the middle of the interval. So our values in 40-49 are all 45, the values in 50-59 are all 55 and so on. This will enable us to estimate the mean, median, and standard deviation.

So with this assumption we have the following grades

frequency

45 3

55 5

65 7

75 6

85 12

95 7

Mean = total of grades / number of grades.

Adding the frequencies gives 40 so number of grades = 40

Total of grades = 45 x 3 + 55 x 5 + 65 x 7 + 75 x 6 + 85 x 12 + 95 x 7 = 3000

So estimated mean = 3000 / 40 = 75

Median. If we list the grades from smallest to largest

a) if the number of grades is odd, the median is the grade in the middle

b) in this case we have two grades in the middle as we have 40 grades so we take the mean of the two middle grades - the 20th and 21st. Looking at the frequency table the 20th grade will be 75 and the 21st grade also 75, so the median will be (75+75) / 2 = 75.

Standard deviation.

To calculate the standard deviation there are 3 steps.

1. Calculate the difference of each grade from the mean, square that, and add these together.

So we do 3 (45-75)²+ 5 (55-75)²+ 7(65-75)²+6(75-75)²+12 (85-75)²+7 (95-75)² giving 3.900 + 5.400 + 7.100 + 6.0 + 12.100 + 7.400 = 9400

Then we divide this by the number of grades, and take the square root √(9400/40) = 15.3

This is the standard deviation.

Mike

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