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:
    • $(x_1+k),~(x_2+k),~(x_3+k),~..,(x_n+k)$ ?
    • $(x_1-k),~(x_2-k),~(x_3-k),~..,(x_n-k)$ ?
    • $kx_1,~kx_2,~kx_3,~..,kx_n$ ?

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

• 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 $

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