ICSE Class 9 Maths Chapter 19 Mean and Median of Ungrouped Data

Mean And Median Of Ungrouped Data

Points To Remember

1. Mean \(\bar{x} = \frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{\Sigma x_i}{n}\)

Where \(\Sigma x_i\) is the sum of observations and \(n\) is the number of observations

or Mean \(\bar{x} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n}\) \(\Rightarrow x_1 + x_2 + x_3 + \ldots x_n = x \times \bar{x}\)

2. Mean for ungrouped data:

(i) Direct method: Let \(x_1, x_2, x_3 \ldots x_n\) be the \(n\) variates and \(f_1, f_2, f_3 \ldots f_n\) be their frequency respectively

Then mean \(\bar{x} = \frac{f_1 x_1 + f_2 x_2 + f_3 x_3 + \ldots + f_n x_n}{f_1 + f_2 + f_3 + \ldots f_n} = \frac{\Sigma f_i x_i}{\Sigma f_i}\)

(ii) Assumed Mean Method: Mean = \(A + \frac{\Sigma f_i d_i}{\Sigma f_i}\)

Where \(A =\) assumed mean, \(f_i\) is the frequency and \(d_i\) is the \((x_i - A)\)

(iii) Step - Deviation Method: Mean \(\bar{x} = A + h + \frac{\Sigma f_i u_i}{\Sigma f_i}\)

where \(= A\) is assumed means \(h = x_2 - x_1\) \(u_i = \frac{x_i - A}{h}\)

3. Median:

(i) Median for ungrouped data: First of all, an angle the observations descending or ascending order

Median = \(\left(\frac{n+1}{2}\right)\) th term where \(n\) is odd observation \(= \frac{1}{2}\left[\frac{n}{2}\text{ th term} + \left(\frac{n}{2}+1\right)\text{ th term}\right]\)

(ii) Median of Discrete Series Method: First arrange the terms in an ascending order or a descending order. Now, prepare a cumulative frequency table. Let the total frequency be \(n\)

(i) If \(n\) is odd then median = \(\frac{n+1}{2}\) th term

If \(n\) is even then median = \(\frac{1}{2}\left[\frac{n}{2}\text{ th term} + \left(\frac{n}{2}+1\right)\text{ th term}\right]\)

Teacher's Note

Understanding mean and median helps us interpret real-world data like average test scores or middle values in datasets, which is essential for making fair comparisons and informed decisions.

Exercise 19(A)

1. The weights of 7 boys in a group are 52 kg 57 kg. 55 kg, 60 kg, 54 kg, 59 kg and 55 kg. Find the mean weight of the group.

Sol. Here \(n = 7\)

Sum of weights of 7 boys is = 52 kg + 57 kg + 55 kg + 60 kg + 54 kg + 59 kg + 55 kg = 392 kg

Mean \(\bar{x} = \frac{\Sigma x_i}{n} = \frac{392}{7} = 56 \text{ kg Ans.}\)

2. The marks obtained by 7 students in a group are 340, 180, 260, 164, 56, 275 and 307 respectively. Find the mean marks per student.

Sol. Here \(n = 7\) and sum of marks of 7 students = 340 + 180 + 260 + 164 + 56 + 275 + 307 = 1582

Mean \(\bar{x} = \frac{\Sigma x_i}{n} = \frac{1582}{7} = 226 \text{ marks Ans.}\)

3. Find the mean of first six prime numbers.

Sol. First 6 prime numbers are 2, 3, 5, 11, 13

Mean \(\bar{x} = \frac{\Sigma x_i}{n} = \frac{2 + 3 + 5 + 7 + 11 + 13}{6} = \frac{41}{6} = 6.83 \text{ Ans.}\)

4. Find the mean of first 10 odd numbers.

Sol. First 10 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Mean \(\bar{x} = \frac{\Sigma x_i}{n} = \frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10} = \frac{100}{10} = 10 \text{ Ans.}\)

5. Find the mean of all the factors of 20.

Sol. Factors of 20 are 1, 2, 4, 5, 10, 20. Here \(n = 6\)

Mean \(\bar{x} = \frac{1 + 2 + 4 + 5 + 10 + 20}{6} = \frac{42}{6} = 7 \text{ Ans.}\)

6. The daily minimum temperature recorded (in degree F), at a place during a week was as under.

Find the mean temperature.

Sol. Here \(n = 6\)

Mean of temperature \(\bar{x} = \frac{\Sigma x_i}{n} = \frac{35.2 + 31.1 + 27.6 + 31.8 + 29.3 + 23.8}{6} = \frac{178.8}{6} = 29.8 \text{ F Ans.}\)

Teacher's Note

Calculating mean temperatures helps meteorologists understand weather patterns and predict seasonal changes, which is useful for planning agricultural activities and daily life.

7. If the mean of 6, 8, 9, \(x\), 13 is 10, find the value of \(x\).

Sol. Mean = 10, Here \(n = 5\)

Mean \(\bar{x} = \frac{\Sigma x_i}{x} = \frac{6 + 8 + 9 + x + 13}{5}\)

\(\Rightarrow 10 = \frac{36 + x}{5} \Rightarrow 36 + x = 50 \Rightarrow x = 50 - 36 = 14 \text{ Ans.}\)

8. The mean of the heights of 6 girls is 148 cm. If the individual heights of five of them are 142 cm, 154 cm, 146 cm, 145 cm and 150 cm, find the height of the sixth girl.

Sol. Mean of height of 6 girls = 148 cm

Total height = 148 \(\times\) 6 = 888 cm

Total heights of 5 girls among them = (142 + 154 + 146 + 145 + 150)cm = 737 cm

Height of the sixth girls = 888 - 737 = 151 cm Ans.

9. The following table shows the weights (in kg) of 15 workers in a factory:

Calculate the mean weight.

Sol. Total number of workers = 15

Mean = \(\frac{\Sigma fx}{\Sigma f} = \frac{975}{15} = 65 \text{ kg Ans.}\)

Teacher's Note

Calculating mean weights in a workplace helps managers understand workforce distribution and make decisions about ergonomic equipment and resource allocation.

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