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# Calculate the Mean, Variance and Standard Deviation of the Following Frequency Distribution. Class:1–1010–2020–3030–4040–5050–60frequency:112918453 - Mathematics

ConceptVariance and Standard Deviation Standard Deviation of a Discrete Frequency Distribution

#### Question

Calculate the mean, variance and standard deviation of the following frequency distribution.

 Class: 1–10 10–20 20–30 30–40 40–50 50–60 Frequency: 11 29 18 4 5 3

#### Solution

Let the assumed mean A = 25.

 Class Mid-Values $\left( x_i \right)$ $d_i = x_i - A$$= x_i - 25$ $d_i^2$ Frequency $\left( f_i \right)$ $f_i d_i$ $f_i d_i^2$ 1–10 5.5 −19.5 380.25 11 −214.5 4182.75 10–20 15 −10 100 29 −290 2900 20–30 25 0 0 18 0 0 30–40 35 10 100 4 40 400 40–50 45 20 400 5 100 2000 50–60 55 30 900 3 90 2700 $\sum_{}f_i =70$ $\sum_{} f_i d_i= −274.5$ $\sum_{} f_i d_i^2=12182.75$
$A + \frac{\sum_{} f_i d_i}{\sum_{} f_i} = 25 + \left( \frac{- 274 . 5}{70} \right) = 25 - 3 . 92 = 21 . 08$

Variance =$\sigma^2 = \left( \frac{1}{N} \sum_{} f_i d_i^2 \right) - \left( \frac{1}{N} \sum_{} f_i d_i \right)^2 = \frac{12182 . 75}{70} - \left( \frac{- 274 . 5}{70} \right)^2 = 174 . 02 - 15 . 37 = 158 . 65$

Standard deviation = $\sigma = \sqrt{158 . 65} = 12 . 6$

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Solution Calculate the Mean, Variance and Standard Deviation of the Following Frequency Distribution. Class:1–1010–2020–3030–4040–5050–60frequency:112918453 Concept: Variance and Standard Deviation - Standard Deviation of a Discrete Frequency Distribution.
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