Question

solve these questions,
find mean, median
and mode.in photo .​

Answer 1

Solution:-

$\boxed{\begin{array}{cccc}\bf Marks&\bf No. \: of &\bf C.F&\bf f_ix_i \\ \bf (x_i) & \bf students\:(f_i) & &\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 15&\sf 2&\sf 2&\sf 30 \\\\\sf 17 &\sf 5&\sf 7&\sf 85 \\\\\sf 20&\sf 10 &\sf 17&\sf 200 \\\\\sf 22&\sf 12&\sf 29&\sf 264\\\\\sf 25&\sf 8 &\sf 37 &\sf 200\\\\\sf 30 &\sf 4 &\sf 41&\sf 120\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{} \\\\\sf &\sf N = \sum f_i = 41&&\sf\sum f_ix_i = 899\end{array}}</p><p></p><p>$

★According to the question:-

★Mean:-

= ∑fi xi / ∑fi

$\sf \: = \frac{899}{41}$

$\sf \: = 21.92$

∴ Mean ↬ 21.92

★Median:-

⟹ $\sf \: Value \: of \: ( \frac{n + 1}{2}) \: th \: \: term$

⟹ $\sf \: Value \: of \: ( \frac{41 + 1}{2})th \: \: term$

⟹ $\sf \: Value \: of \: 21st \: term \: = 22$

∴ Median ↬ 22

★Mode:-(Empirical formula)

$\sf \: Mode \: \: = 3(median) \: - 2(mean)$

$\sf \: = 3 \times 22 - 2 \times 21.92$

$\sf \: = 66 - 43.84$

$\sf \: = 22.16$

∴ Mode ↬22.16

Therefore,

• Mean - 21.92

• Median - 22

• Mode - 22.16