Statistics is the branch of mathematics that deals with collecting, organizing, analyzing, interpreting, and presenting data. It helps us make sense of complex data and draw meaningful conclusions for decision-making or predictions.
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Important Statistics formulas
- Class Interval = Upper Limit - Lower Limit
Mean
(\bar x) For Ungrouped Data
- Mean
( \bar x) = \frac{x_1 + x_2 + x_3 + .....x_n}{n} Mean
(\bar x) For Grouped Data
- Direct Method :
( \bar x) = \frac{\sum f_i \ x_i}{\sum f_i}
Where, fi = frequency of each observation/class,
xi = value of observations or midpoint for each class interval.
- Assumed Mean Method :
( \bar x) = A + \frac{\sum f_i \ d_i}{\sum f_i}
Where, fi = frequency of each observation/class,
di = deviation of each observations or midpoint of each class interval.( d_i = x_i - A )
- Step- Deviation Method :
( \bar x) = A + \frac{\sum f_i \ u_i}{\sum f_i}. \ h
Where, fi = frequency of each observation/class,
ui = Step deviation of each observations or midpoint of each class interval.( d_i = x_i - A ) Median For Ungrouped Data
- For odd number of Observations : Median =
[\frac{(n + 1)}{2}]^{th} \ term
Where, n= number of observations.
- For even number of observations : Median =
\frac{[(\frac{n}{2})^{th}\ term + \{ \frac{n}{2} + 1 \}^{th} \ term]}{2}
Where, n= number of observations.Median For Grouped Data
- Median =
l + [\frac{(\frac{n}{2}- cf)}{f}] \times h
Where, l = Lower Limit of Median Class
n = Number of Observations
f = Frequency of Median Class
h = Class Size
cf = Cumulative Frequency of Class Preceding Median ClassMode for Ungrouped Data
- Mode = Observation with highest frequency / number of occurrence.
Mode for Grouped Data
- Mode =
l + \{\frac{f_1 - f_2}{2f_1 - f_0-f_2} \} \times h
Where, l = lower limit of the modal class.
h = size of the class interval,
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class, and
f2 = frequency of the class succeeding the modal class.Relation between Mean Median and Mode
- Mode = 3 Median – 2 Mean
Statistics Practice Questions - Solved
Question 1: If the marks scored by the students in a class test out of 50 are,
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Number of Students | 5 | 8 | 9 | 6 | 3 |
a) What is the class interval of 20 - 30 ?
b) How many students scored in the range of 20 - 40?
Solution:
a) Class interval = Upper limit - Lower limit
Class interval of 20-30 = 30 - 20 = 10b) The number of students scored in range of 20 - 40 = frequency of class 20 - 30 + frequency of class 30 - 40
The number of students scored in range of 20 - 40 = 9 + 6 = 15
Question 2: Calculate the average of the following numbers: 6, 8, 2, 3, 12, 14.
Solution:
Mean = Sum of all given observations/ Total number of observations
Total number of observations = 6
Mean = 6 + 8 + 2 + 3 + 12 + 14 / 6 = 45/6 = 7.5Thus, the Mean of 6, 8, 2, 3, 12, 14 is 7.5
Question 3: Find the mean of the first 10 even numbers.
Solution:
Sum of first n even numbers = n × (n + 1)
Sum of first 10 even numbers = 10 × ( 10 + 1) = 10 × 11 = 110
Mean of first 10 even numbers = 110 / 10 = 11Thus, the mean of first 10 even numbers is 11.
Questions 4 : The following are the monthly salaries (in $1000s) of 15 employees in a company:
30, 45, 25, 35, 40, 50, 60, 55, 45, 50, 25, 35, 40, 45, 50.
Find the median salary.
Solution:
Sort the data: [25, 25, 30, 35, 35, 40, 40, 45, 45, 45, 50, 50, 50, 55, 60].
Total number of elements (N) = 15 (odd).
Median =\frac{N}{2}^{th}\ term = 45.
Question 5 : The following table represents the marks obtained by students in a test. Find the median of the marks.
| Class Interval | Frequency |
|---|---|
| 0–20 | 5 |
| 20–40 | 8 |
| 40–60 | 12 |
| 60–80 | 15 |
| 80–100 | 10 |
Solution:
Calculate the cumulative frequency:
Class Interval Frequency Cumulative Frequency 0–20 5 5 20–40 8 13 40–60 12 25 60–80 15 40 80–100 10 50 Total frequency (N) = 50.
Median class = class where
\frac{N}{2}^{th} observation\frac{50}{2} = 25 exists.
Median class = 40 – 60.Median =
L + \left( \frac{\frac{N}{2} - F}{f} \right) \cdot h Where:
L = 40 (lower boundary of the median class),
N=50 ( total number of observations )
F = 13 ( cumulative frequency before the median class ),
f = 12 (frequency of the median class),
h=20 (class width).Substituting the values:
Median =40 + \left( \frac{25 - 13}{12} \right) \cdot 20
Median =40 + \left( \frac{12}{12} \right) \cdot 20 = 40 + 20 = 60
Question 6: The following table shows the marks scored by students in a test. Calculate the mean using the assumed mean method.
| Class Interval | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 8 |
| 30–40 | 12 |
| 40–50 | 15 |
| 50–60 | 10 |
Solution:
Choose A=35 (assumed mean).
Calculate di = xi − A , where xi​ is the class midpoint.
Class Interval Frequency (fi​) Midpoint (xi​) di=xi − A fi⋅di 10–20 5 15 −20 −100 20–30 8 25 −10 −80 30–40 12 35 0 0 40–50 15 45 10 150 50–60 10 55 20 200
- ∑fi​ = 50
- ∑fidi = 170
For assumed mean method :
Mean
( \bar x) = A + \frac{\sum f_i \ d_i}{\sum f_i} Mean =
35 + \frac{170}{50} = 35 + 3.4 = 38.4
Question 7 :The following table shows the daily wages of workers in a factory. Use the assumed mean method to calculate the mean daily wage.
| Wages (in ₹) | Frequency |
|---|---|
| 50–60 | 6 |
| 60–70 | 10 |
| 70–80 | 20 |
| 80–90 | 12 |
| 90–100 | 8 |
Solution:
Assume A = 75 (from the midpoint of the central class, 70–80).
Find the midpoints (xi):
Class Interval Frequency (fi) Midpoint (xi​) di=xi−A fi⋅di 50–60 6 55 −20-20−20 −120-120−120 60–70 10 65 −10-10−10 −100-100−100 70–80 20 75 000 000 80–90 12 85 101010 120120120 90–100 8 95 202020 160160160 Apply the formula for the mean:
Mean =
A + \frac{\sum f_i d_i}{\sum f_i} Substitute the values:
- ∑fi = 56
- ∑fidi=60
Mean =
75 + \frac{60}{56} ​ = 75 + 1.07 = 76.07
Question 8 : Find the mode of the following dataset representing the number of hours studied by students in a week:
6, 8, 9, 6, 7, 8, 6, 9, 7, 8, 8, 6, 7, 8, 9
Solution:
Organize the data and count the frequency of each value:
Hours Studied Frequency 6 4 7 3 8 5 9 3 Identify the mode:
The mode is the value with the highest frequency.
Mode = 8 (Frequency: 5)
Question 9: Find the mode for the following frequency distribution of students' marks:
| Marks Interval | Frequency |
|---|---|
| 10–20 | 3 |
| 20–30 | 7 |
| 30–40 | 12 |
| 40–50 | 18 |
| 50–60 | 10 |
| 60–70 | 5 |
Solution:
Identify the modal class: The class with the highest frequency is 40–5040–5040–50, so it is the modal class.
Mode =
L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \cdot h Where:
- L = 40 (lower boundary of the modal class),
- fm = 18 (frequency of the modal class),
- f1 = 12 (frequency of the class before modal class),
- f2 = 10 (frequency of the class after modal class),
- h=10 (class width).
Mode =
40 + \left( \frac{18 - 12}{2(18) - 12 - 10} \right) \cdot 10 Mode =
40 + \left( \frac{6}{36 - 22} \right) \cdot 10 Mode =
40 + \left( \frac{6}{14} \right) \cdot 10 Mode = 40 + 4.29 = 44.29
Question 10: A dataset is positively skewed. If the mode is 40 and the median is 45, find the mean using the relationship between mean, median, and mode.
Solution:
Using the formula: Mode = 3(Median) − 2(Mean)
Mode = 3(45)−2(40)= 135 − 80 = 55
Unsolved Practice Question on Statistics
Question 1: The following table represents the heights (in cm) of a group of students:
| Height (cm) | Frequency |
|---|---|
| 140–150 | 5 |
| 150–160 | 12 |
| 160–170 | 8 |
| 170–180 | 10 |
Find the class width of the intervals.
Question 2: Given the following raw data:
35, 40, 42, 45, 50, 52, 55, 60, 65, 68, 70, 72
Find the frequency of the values ranging from 40 to 60.
Question 3: The following data represents the marks scored by students in a test:
| Marks Interval | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 7 |
| 30–40 | 8 |
| 40–50 | 10 |
| 50–60 | 5 |
Calculate the mean using the direct method.
Question 4 : The wages of workers in a factory are shown below. Use the assumed mean method to calculate the mean wage:
| Wages (₹) | Frequency |
|---|---|
| 200–300 | 6 |
| 300–400 | 8 |
| 400–500 | 15 |
| 500–600 | 10 |
| 600–700 | 5 |
Question 5. The following data represents the monthly expenses (in $) of 12 families: 450, 500, 550, 600, 620, 700, 720, 750, 800, 850, 900, 950. Find the median of the dataset.
Question 6: The table below shows the marks obtained by students in a class:
| Marks Interval | Frequency |
|---|---|
| 10–20 | 4 |
| 20–30 | 8 |
| 30–40 | 10 |
| 40–50 | 15 |
| 50–60 | 12 |
Find the mode using the formula for grouped data.
Question 7 : The following table represents the number of hours students spent studying:
| Hours Studied | Frequency |
|---|---|
| 0–2 | 4 |
| 2–4 | 6 |
| 4–6 | 10 |
| 6–8 | 8 |
| 8–10 | 5 |
Find the median using the formula for grouped data.
Question 8: The following dataset represents the number of books borrowed by students in a week: 2, 3, 5, 3, 4, 3, 2, 5, 3, 2, 4, 5, 3, 3, 4. Find the mode of the dataset.
Question 9: The mean and median of a dataset are given as 25 and 22, respectively. Use the relationship between mean, median, and mode to calculate the mode.
Question 10: Given the frequency distribution below, answer the following questions:
- a) What is the total frequency?
- b) Identify the modal class.
- c) Calculate the median class.
| Class Interval | Frequency |
|---|---|
| 0–10 | 6 |
| 10–20 | 12 |
| 20–30 | 18 |
| 30–40 | 14 |
| 40–50 | 10 |
Answer key for Unsolved Questions:
- 10
- 7
- 35.86
- 450
- 710
- 46.25
- 5.3
- 3
- 16
- a) 60
b) 20-30
c) 30-40