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Statistics - Mean of Grouped Data

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Grouped data consists of data points organized into class intervals. To calculate the mean, we first find the representative value for each class, called the Class Mark (xix_i).

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The Class Mark (xix_i) is the midpoint of the class interval: xi=Upper Class Limit+Lower Class Limit2x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}.

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The Direct Method is used when the values of class marks xix_i and frequencies fif_i are small, making their product easy to calculate.

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The Assumed Mean Method simplifies calculations by shifting the origin to a value aa (usually the middle class mark). We calculate deviations di=xi−ad_i = x_i - a.

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The Step-Deviation Method further scales down the deviations by the class size hh, where ui=xi−ahu_i = \frac{x_i - a}{h}. This is particularly helpful when class sizes are uniform and xix_i values are large.

📐Formulae

Class Mark (xi)=Upper Limit+Lower Limit2\text{Class Mark } (x_i) = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}

Mean (xˉ)=∑fixi∑fi (Direct Method)\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \text{ (Direct Method)}

Mean (xˉ)=a+∑fidi∑fi (Assumed Mean Method, where di=xi−a)\text{Mean } (\bar{x}) = a + \frac{\sum f_i d_i}{\sum f_i} \text{ (Assumed Mean Method, where } d_i = x_i - a)

Mean (xˉ)=a+(∑fiui∑fi)×h (Step-Deviation Method, where ui=xi−ah)\text{Mean } (\bar{x}) = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \text{ (Step-Deviation Method, where } u_i = \frac{x_i - a}{h})

💡Examples

Problem 1:

Find the mean of the following distribution showing the marks of 30 students:

Class Interval10−2525−4040−5555−7070−8585−100Frequency (fi)237666\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Class Interval} & 10-25 & 25-40 & 40-55 & 55-70 & 70-85 & 85-100 \\ \hline \text{Frequency } (f_i) & 2 & 3 & 7 & 6 & 6 & 6 \\ \hline \end{array}

Solution:

  1. Find Class Marks (xix_i):
  • For 10−2510-25, xi=10+252=17.5x_i = \frac{10+25}{2} = 17.5
  • For 25−4025-40, xi=25+402=32.5x_i = \frac{25+40}{2} = 32.5
  • Similarly, other class marks are 47.5,62.5,77.5,92.547.5, 62.5, 77.5, 92.5.
  1. Calculate fixif_i x_i:
  • 2×17.5=35.02 \times 17.5 = 35.0
  • 3×32.5=97.53 \times 32.5 = 97.5
  • 7×47.5=332.57 \times 47.5 = 332.5
  • 6×62.5=375.06 \times 62.5 = 375.0
  • 6×77.5=465.06 \times 77.5 = 465.0
  • 6×92.5=555.06 \times 92.5 = 555.0
  1. Summation:
  • ∑fi=2+3+7+6+6+6=30\sum f_i = 2+3+7+6+6+6 = 30
  • ∑fixi=35.0+97.5+332.5+375.0+465.0+555.0=1860.0\sum f_i x_i = 35.0 + 97.5 + 332.5 + 375.0 + 465.0 + 555.0 = 1860.0
  1. Apply Mean Formula: xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i} xˉ=186030\bar{x} = \frac{1860}{30} xˉ=62\bar{x} = 62

Explanation:

We use the Direct Method here. First, we find the mid-point of each class interval to represent the class. Then, we multiply each midpoint by its corresponding frequency. The sum of these products divided by the total number of students gives the average marks.

Problem 2:

Find the mean of the following data using the Assumed Mean Method, taking a=50a = 50:

Class0−2020−4040−6060−8080−100fi581278\begin{array}{|c|c|c|c|c|c|} \hline \text{Class} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 \\ \hline f_i & 5 & 8 & 12 & 7 & 8 \\ \hline \end{array}

Solution:

  1. Class Marks (xix_i): 10,30,50,70,9010, 30, 50, 70, 90.
  2. Assume a=50a = 50. Calculate deviations di=xi−50d_i = x_i - 50:
  • d1=10−50=−40d_1 = 10 - 50 = -40
  • d2=30−50=−20d_2 = 30 - 50 = -20
  • d3=50−50=0d_3 = 50 - 50 = 0
  • d4=70−50=20d_4 = 70 - 50 = 20
  • d5=90−50=40d_5 = 90 - 50 = 40
  1. Calculate fidif_i d_i:
  • 5×(−40)=−2005 \times (-40) = -200
  • 8×(−20)=−1608 \times (-20) = -160
  • 12×0=012 \times 0 = 0
  • 7×20=1407 \times 20 = 140
  • 8×40=3208 \times 40 = 320
  1. Summation:
  • ∑fi=40\sum f_i = 40
  • ∑fidi=−200−160+0+140+320=100\sum f_i d_i = -200 - 160 + 0 + 140 + 320 = 100
  1. Apply Formula: xˉ=a+∑fidi∑fi\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} xˉ=50+10040\bar{x} = 50 + \frac{100}{40} xˉ=50+2.5=52.5\bar{x} = 50 + 2.5 = 52.5

Explanation:

The Assumed Mean Method reduces the size of the numbers we multiply. By subtracting a=50a = 50 from every class mark, we work with smaller deviations did_i. The final adjustment ∑fidi∑fi\frac{\sum f_i d_i}{\sum f_i} is added back to the assumed mean to find the actual mean.