Page 286 - Start Up Mathematics_7

(e) Find the number of days for which the temperature recorded was below the

mean temperature.
Solution: Let’s arrange the data in ascending order:

14, 16, 18, 21, 22, 23, 25, 25, 27, 29
(a) The maximum temperature reached on the hottest day is 29°C.

(b) The least of the maximum day temperatures is 14°C.
(c) Range of the data = 29 – 14 = 15°C

(d) Mean = 14 + 16 + 18 + 21+ 22 + 23 + 25 + 25 + 27 + 29
10
220
= = 22°C
70
(e) The temperature was below the mean temperature for 4 days.
Example 4: The time (in minutes) by which Neeraj reached late to school is recorded in the
following table.

Monday Tuesday Wednesday Thursday Friday Saturday
2 4 1 0 2 45
Find the mean or average time by which he is late.

Solution: Mean = Sum of all observations
Total number of observations
4
1 0
2
= 2 + + + + + 45 = 54
6 6
= 9 minutes
Except for one day, Neeraj is not late to his school by more than 4 minutes. The
mean or average late timing of 9 minutes is in fact more than double of 4 minutes.
This is because a single extreme value of 45 minutes has distorted the value of the
mean.

Mean is highly affected by extreme values.

Arithmetic mean of grouped data
In the previous section we have learnt to find the arithmetic mean of series of individual observations
also called raw data. If the data consists of a large number of observations, we know it can be

converted into frequency distribution. In a frequency table, we have observations x , x , x , ..., x
2
1
n
3
with corresponding frequencies f , f , f , ..., f .
2
3
n
1
Σ f xi
i
The arithmetic mean of these values is defined as ,
Σ f i
f x 1 + f x 2 + f x 3 + ... + f x n
1
3
2
n
i.e., Mean =
f 1 + f 2 + f 3 + ... + f n
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