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Statistics - Mean of Grouped Data (Direct and Assumed Mean Method)

Grade 10CBSE

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

πŸ”‘Concepts

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Grouped Data and Class Intervals: In statistics, large sets of data are organized into groups called class intervals (e.g., 10βˆ’2010-20, 20βˆ’3020-30). Visually, these intervals represent the width of bars in a histogram, where the 'Lower Class Limit' is the left boundary and the 'Upper Class Limit' is the right boundary of each group.

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Class Mark (xix_i): The class mark is the midpoint of a class interval, representing the entire group for calculation purposes. Visually, if you imagine a number line, the class mark is the exact center point between the upper and lower limits of an interval. It is calculated as the average of the two limits.

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Frequency (fif_i): This represents the number of observations falling within a specific class interval. In a visual frequency distribution table, this is the count associated with each row, determining the 'weight' or importance of that interval in the overall mean.

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The Direct Method: This is the basic approach to finding the mean where we multiply each class mark (xix_i) by its corresponding frequency (fif_i) and divide the total sum by the total number of observations. Visually, this treats the class mark as the representative 'height' of data for that group.

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Assumed Mean Method: To simplify calculations when class marks are large, we select one class mark (usually the middle one) as the 'Assumed Mean' (aa). Visually, this is like shifting the origin of our data set to aa, calculating how far other points 'deviate' from this center point.

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Deviation (did_i): In the Assumed Mean Method, the deviation did_i is the difference between each class mark and the assumed mean (di=xiβˆ’ad_i = x_i - a). Visually, positive deviations represent values to the right of the assumed mean, while negative deviations represent values to the left.

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Sigma Notation (βˆ‘\sum): This mathematical symbol represents the sum of a sequence of numbers. In a table, βˆ‘fi\sum f_i represents the total of the frequency column, while βˆ‘fixi\sum f_i x_i represents the total of the product column, serving as the aggregate value of all data points.

πŸ“Formulae

Class Mark: xi=UpperΒ ClassΒ Limit+LowerΒ ClassΒ Limit2x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}

Mean (Direct Method): xΛ‰=βˆ‘fixiβˆ‘fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Deviation: di=xiβˆ’ad_i = x_i - a (where aa is the Assumed Mean)

Mean (Assumed Mean Method): xΛ‰=a+βˆ‘fidiβˆ‘fi\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

πŸ’‘Examples

Problem 1:

Calculate the mean for the following distribution using the Direct Method: Class Interval: 0βˆ’10,10βˆ’20,20βˆ’30,30βˆ’400-10, 10-20, 20-30, 30-40 Frequency: 7,8,12,137, 8, 12, 13

Solution:

  1. Find Class Marks (xix_i):
  • For 0βˆ’10:x1=0+102=50-10: x_1 = \frac{0+10}{2} = 5
  • For 10βˆ’20:x2=10+202=1510-20: x_2 = \frac{10+20}{2} = 15
  • For 20βˆ’30:x3=20+302=2520-30: x_3 = \frac{20+30}{2} = 25
  • For 30βˆ’40:x4=30+402=3530-40: x_4 = \frac{30+40}{2} = 35
  1. Multiply fif_i and xix_i:
  • f1x1=7Γ—5=35f_1x_1 = 7 \times 5 = 35
  • f2x2=8Γ—15=120f_2x_2 = 8 \times 15 = 120
  • f3x3=12Γ—25=300f_3x_3 = 12 \times 25 = 300
  • f4x4=13Γ—35=455f_4x_4 = 13 \times 35 = 455
  1. Calculate Sums:
  • βˆ‘fi=7+8+12+13=40\sum f_i = 7 + 8 + 12 + 13 = 40
  • βˆ‘fixi=35+120+300+455=910\sum f_ix_i = 35 + 120 + 300 + 455 = 910
  1. Calculate Mean:
  • xΛ‰=βˆ‘fixiβˆ‘fi=91040=22.75\bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{910}{40} = 22.75

Explanation:

This approach uses the Direct Method. We first find the midpoints of each interval to represent the data, then find the weighted total by multiplying by frequencies, and finally divide by the total count.

Problem 2:

Find the mean of the following data using the Assumed Mean Method: Class: 100βˆ’120,120βˆ’140,140βˆ’160,160βˆ’180,180βˆ’200100-120, 120-140, 140-160, 160-180, 180-200 Frequency: 12,14,8,6,1012, 14, 8, 6, 10

Solution:

  1. Find Class Marks (xix_i): 110,130,150,170,190110, 130, 150, 170, 190.
  2. Choose Assumed Mean (aa): Let a=150a = 150 (the middle value).
  3. Calculate Deviations (di=xiβˆ’150d_i = x_i - 150):
  • d1=110βˆ’150=βˆ’40d_1 = 110 - 150 = -40
  • d2=130βˆ’150=βˆ’20d_2 = 130 - 150 = -20
  • d3=150βˆ’150=0d_3 = 150 - 150 = 0
  • d4=170βˆ’150=20d_4 = 170 - 150 = 20
  • d5=190βˆ’150=40d_5 = 190 - 150 = 40
  1. Calculate fidif_id_i:
  • 12Γ—(βˆ’40)=βˆ’48012 \times (-40) = -480
  • 14Γ—(βˆ’20)=βˆ’28014 \times (-20) = -280
  • 8Γ—0=08 \times 0 = 0
  • 6Γ—20=1206 \times 20 = 120
  • 10Γ—40=40010 \times 40 = 400
  1. Calculate Sums:
  • βˆ‘fi=12+14+8+6+10=50\sum f_i = 12 + 14 + 8 + 6 + 10 = 50
  • βˆ‘fidi=βˆ’480βˆ’280+0+120+400=βˆ’240\sum f_id_i = -480 - 280 + 0 + 120 + 400 = -240
  1. Calculate Mean:
  • xΛ‰=a+βˆ‘fidiβˆ‘fi=150+βˆ’24050\bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 150 + \frac{-240}{50}
  • xΛ‰=150βˆ’4.8=145.2\bar{x} = 150 - 4.8 = 145.2

Explanation:

The Assumed Mean Method reduces the size of the numbers we work with. By picking 150150 as a center, we only deal with small deviations, making the multiplication and addition much easier before adding the correction back to the assumed mean.