Mean Median Mode
Published:
This post covers Introduction to descriptive statistics.
Questions
Mean
$\bar{x} = \frac{\Sigma x_if_i}{\Sigma f_i} $
Raw Data
- A cricketer’s scores in five ODI matches are as follows: $12, 34, 45, 50, 24$. Find the mean score.
- $\bar{x} = 33$
- If the heights of 5 people are $142 cm, 150 cm, 149 cm, 156 cm, and 153 cm$. Find the mean height.
- $\bar{x} = 150$
Tabular (Frequency Distribution) Form
Find the mean of the following distribution:
x 4 6 9 10 15 f 5 10 10 7 8 $\bar{x} = \frac{\Sigma x_if_i}{\Sigma f_i} = 9$
The following table indicates the data on the number of patients visiting a hospital in a month. Find the average number of patients visiting the hospital in a day.
Number of Patients Number of days visiting hospital 0-10 2 10-20 6 20-30 9 30-40 7 40-50 4 50-60 2 $class~mark = x_i = \frac{lower+upper}{2}$
$\bar{x} = \frac{\Sigma x_if_i}{\Sigma f_i} = 28.67$
Challenging:
- Let the mean of $x_1,~ x_2,~ x_3,~…,~x_n$ be $A$, then what is the mean of:
- $(x_1+k),~(x_2+k),~(x_3+k),~..,(x_n+k)$ ?
- $(x_1-k),~(x_2-k),~(x_3-k),~..,(x_n-k)$ ?
- $kx_1,~kx_2,~kx_3,~..,kx_n$ ?
- Let the mean of $x_1,~ x_2,~ x_3,~…,~x_n$ be $A$, then what is the mean of:
Median
Middlemost value after arranging the data in order
Ungrouped data
Arrange the data in ascending/descending order
$Median = \frac{n+1}{2}th$ observation, if $n$ is odd
$Median = \frac{\frac{n}{2}th~obs+(\frac{n}{2}+1)th~obs}{2}th$ observation, if $n$ is even
- Find the median of $56, 67, 54, 34, 78, 43, 23$
- Median = $54$
- Find the median of $50, 67, 24, 34, 78, 43$.
- Median = $46.5$
Grouped Data
$n = \Sigma f_i$
$ Median = l + [\frac{\frac{n}{2}-c}{f}] * h $
- $l =$ lower limit
- $c =$ cumulative frequency of the class preceding the median class
- $f = $ frequency of the median class
- $h = $ class size
Find the median for
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
---|---|---|---|---|---|
Frequency | 2 | 12 | 22 | 8 | 6 |
Step 1: Compute Cumulative Frequencies for the median
Classes Number of students Cumulative frequency 0-10 2 2 10-20 12 2 + 12 = 14 20-30 22 14 + 22 = 36 30-40 8 36 + 8 = 44 40-50 6 44 + 6 = 50 Step 2:
- $n = 50$
- Median class is $\frac{n}{2} = 25 \implies 20-30 ~(class)$
- $l = 20; f = 22; c=14; h = 10$
- $ Median = l + [\frac{\frac{n}{2}-c}{f}] * h = 20 + \frac{25-14}{22} * 10 = 25$
Mode
Ungrouped Data
- What is the mode of $6, 8, 9, 3, 4, 6, 7, 6, 3$
- Mode = $6$
- What is the mode of A = {$1, 2, 3, 3, 4, 4, 5, 6$}
- Mode[A] = {$3, 4$}
- Bimodal
- What is the mode of B = {$1, 2, 3, 3, 4, 4, 5, 5, 6$}
- Mode[B] = {$3, 4, 5$}
- Trimodal
Grouped Data
For continuous data
Step 1: Find modal class i.e. class with maximum frequency
Step 2: Find mode:
Mode = $l + \Big[ \frac{f_m - f_1}{2f_m - f_1 - f_2} \Big] * h$
where, $l =$ lower limit of modal class,
$f_m =$ frequency of modal class,
$f1 =$ frequency of class preceding modal class,
$f2 =$ frequency of class succeeding modal class,
$h =$ class width
Find mode of:
Marks Obtained 0-20 20-40 40-60 60-80 80-100 Number of students 5 10 12 6 3 - Highest Frequency = 12; Thus, modal class is $40-60$
- $l= 40; f_m = 12; f_1=10; f_2=6; h = 20$
Mode = $40 + \Big[ \frac{12 - 10}{2*12 - 10 - 6} \Big] * 20 = 45$
Calculate Mean, Median, Mode
- Instructions: Click on the green board and observe the mean, median, and mode of the numbers appearing on the screen.
Source: https://www.cuemath.com/data/mean-median-mode/
Empirical Relationship
$2Mean + Mode =3Median$
Q1: Find the missing frequency $p$ when mean is $20.6$
x 10 15 20 25 35 f 3 10 ==p== 7 5 - $20.6 = \frac{530+20p}{25+p} \implies p = 25$
Q2: The mean of $5$ numbers is $18$. If one number is excluded, their mean is $16$ Find the excluded number.
- $\bar{x} = 18;~ n = 5 \implies \Sigma x = 90$
- $\frac{90 - a}{4} = 16 \implies a = 26$
A survey on the heights (in cm) of 50 girls of class X was conducted at a school and the following data was obtained as below. Find the mode and median.
| Height (in cm) | 120-130 | 130-140 | 140-150 | 150-160 | 160-170 | Total | | :—————–: | ———– | ———– | ———– | ———– | ———– | ——— | | Number of girls | 2 | 8 | 12 | 20 | 8 | 50 |
Mode
- Modal Class: $150-160$
- $l=150,~ h=10,~ f_m=20,~ f_1=12,~ f_2=8$
- $Mode = l + \Big[ \frac{f_m - f_1}{2f_m - f_1 - f_2} \Big] * h$
- $Mode = 150 + \Big[ \frac{20 - 12}{2*20 - 12 - 8} \Big] * 10 = 154$
Median
Class Intervals No. of girls (fi) Cumulative frequency (c) 120-130 2 2 130-140 8 2+8=10 140-150 12 = f1 10+12=22 (c) 150-160 20 = fm 22+20=42 160-170 8 = f2 42+8=50 (n) - $n = 50 \implies median = 25 \implies Median~class=150-160$
- $l=150,~ c=22,~ f=20,~ h=10$
- $ Median = l + [\frac{\frac{n}{2}-c}{f}] * h $
- $ Median = 150 + [\frac{\frac{50}{2} - 22}{20}] * 10 = 151.5 $
- $n = 50 \implies median = 25 \implies Median~class=150-160$
Source: https://www.cuemath.com/data/mean-median-mode/ |
Difference between Mean and Median
A department of an organization has 5 employees which include a supervisor and four executives. The executives draw a salary of 10,000 per month while the supervisor gets 40,000.
- $Mean = 16,000$
- $Median = 10,000 $
- Mean salary of 16,000 does not give even an estimated salary of any of the employees whereas the median salary represents the data more effectively.
- One of the weaknesses of mean is that it gets affected by extreme values.
How extreme values affect mean and median
Source: https://www.cuemath.com/data/mean-median-mode/ Right-skewed: Mean to the right of the Median; Tail is on the right
- Left-skewed: Mean is to the left of the Median; Tail is on the left