Q74 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

$\begin{aligned} &\text { Mean deviation about the mean for the following data is }\\ &\begin{array}{|l|c|c|c|c|c|} \hline \text { Class Interval } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \hline \text { Frequency } & 1 & 2 & 3 & 2 & 1 \\ \hline \end{array} \end{aligned}$

  • A
    5
  • Correct answer
    B
    $\frac{16}{3}$
  • C
    6
  • D
    $\frac{19}{3}$

Solution

Step 1: Find the class midpoints \(\left(x_i\right)\) : The midpoint for a class interval is calculated as: \(x_i=\frac{\text { Lower limit }+ \text { Upper limit }}{2}\) - \(x_1=\frac{0+6}{2}=3\) - \(x_2=\frac{6+12}{2}=9\) - \(x_3=\frac{12+18}{2}=15\) - \(x_4=\frac{18+24}{2}=21\) - \(x_5=\frac{24+30}{2}=27\) Thus, \(x_i=3,9,15,21,27\). Step 2: Compute the mean \((\bar{x}):\) The mean is calculated as: \(\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\) - \(\sum f_i=1+2+3+2+1=9\) - \(\sum f_i x_i=(1 \cdot 3)+(2 \cdot 9)+(3 \cdot 15)+(2 \cdot 21)+(1 \cdot 27)\) \(\begin{gathered} \sum f_i x_i=3+18+45+42+27=135 \\ \bar{x}=\frac{135}{9}=15 \end{gathered}\) Step 3: Find \(\left|x_i-\bar{x}\right|\) and \(f_i\left|x_i-\bar{x}\right|\) : \(\left|x_i-\bar{x}\right|=\text { Absolute deviation from the mean. }\) - For \(x_1=3:|3-15|=12, f_1\left|x_i-\bar{x}\right|=1 \cdot 12=12\) - For \(x_2=9:|9-15|=6, f_2\left|x_i-\bar{x}\right|=2 \cdot 6=12\) - For \(x_3=15:|15-15|=0, f_3\left|x_i-\bar{x}\right|=3 \cdot 0=0\) - For \(x_4=21:|21-15|=6, f_4\left|x_i-\bar{x}\right|=2 \cdot 6=12\) - For \(x_5=27:|27-15|=12, f_5\left|x_i-\bar{x}\right|=1 \cdot 12=12\) Step 4: Calculate mean deviation about the mean: \(\begin{gathered} \text {Mean deviation }=\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i} \\ \sum f_i\left|x_i-\bar{x}\right|=12+12+0+12+12=48 \\ \text {Mean deviation }=\frac{48}{9}=\frac{16}{3} \end{gathered}\) Final Answer: The mean deviation about the mean is \(\frac{16}{3}\) (Option 2).

Hence, the correct answer is option: 'B'.

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