Question

find the mean median and mode

Answer 1

mean = sum of all observation / no. of observation
mode = hightest no. of occurance
median = (odd ) n and plus one upon two

Answer 2

ANSWER↓

Mean    =  42.2

Median = 32.5

Mode    = 13.1

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STEP BY STEP EXPLANATION ↓

$\begin{array}{| c | c | c | c | c |}\cline{1-5} \bold{C.I}&\bold{Fq\;(f)}&\bold{CM\:(x)}&\bold{Cf}&\bold{fx} \\ \cline{1-5} 10-20&4&15&4&60 \\ \cline{1-5} 20-30&8&25&12&200\\ \cline{1-5} 30-40&10&35&22&350\\ \cline{1-5} 40-50&12&45&34&540\\ \cline{1-5} 50-60&10&65&44&550\\ \cline{1-5} 60-70&4&55&48&260\\ \cline{1-5} 70-80&2&75&50&150\\ \cline{1-5} &\bold{\sum f=50}&&&\bold{\sum fx=2110} \\ \cline{1-5}\end{array}$

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Formula's which is used in this question ↓

$\boxed{\implies\:Mean=\dfrac{\sum fx}{\sum f}}$

$\boxed{\implies\:Median=l+h\:\dfrac{N/2-Cf}{f}}$

$\boxed{\implies\:Mode=3\times Median-2\times Mean}$

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MEAN OF THE DATA ↓

$\implies\:\overline{x}=\dfrac{\sum fx}{\sum f}\\\\\\\implies\:\overline{x}=\dfrac{2110}{50}\\\\\\\implies\:\overline{x}=42.2$

$\huge{\boxed{\therefore\:Mean=42.2}$

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MEDIAN OF THE DATA ↓

Here,

• Lower Limit [l] = 40

• Height [h] = Upper limit - Lower Limit = 10

• N = 50 ⇒ N/2 = 25

• Cumulative Frequency [Cf] =34

• Frequency [f] = 12

$\boxed{\implies\:Median=l+h\:\dfrac{N/2-Cf}{f}}\\\\\\\implies\:Median=40+10\:\dfrac{25-34}{12}\\\\\\\implies\:Median=40+5\:\dfrac{-9}{6}\\\\\\\implies\:Median=32.5$

$\huge{\boxed{\therefore\:Median=32.5}}$

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MODE OF THE GIVEN DATA ↓

⇒ Mode = 3×Median - 2×Mean

⇒ Mode = 3×32.5 - 2×42.2

⇒ Mode = 97.5 - 84.4

⇒ Mode = 13.1

$\huge{\boxed{\therefore\:Mode=13.1}}$