Find the mean deviation about the mean for the following data: Marks obtained: $10 ext{-}20$, $20 ext{-}30$, $30 ext{-}40$, $40 ext{-}50$, $50 ext{-}60$, $60 ext{-}70$, $70 ext{-}80$; Number of Students: $2$, $3$, $8$, $14$, $8$, $3$, $2$.

Question

Find the mean deviation about the mean for the following data: Marks obtained: $10	ext{-}20$, $20	ext{-}30$, $30	ext{-}40$, $40	ext{-}50$, $50	ext{-}60$, $60	ext{-}70$, $70	ext{-}80$; Number of Students: $2$, $3$, $8$, $14$, $8$, $3$, $2$.

Accepted Answer

>

> [!TLDR]
> The mean of the grouped data is $45$; the mean deviation about the mean, computed from $\sum f_i|x_i-\bar x|/N$, comes out to $10$.
>
> **Step 1 — Midpoints ($x_i$) of each class:**
> $10$–$20	o15$, $20$–$30	o25$, $30$–$40	o35$, $40$–$50	o45$, $50$–$60	o55$, $60$–$70	o65$, $70$–$80	o75$.
>
> **Step 2 — Compute $\sum f_ix_i$ and $N=\sum f_i$:**
>
> | $x_i$ | $f_i$ | $f_ix_i$ |
> |---|---|---|
> | 15 | 2 | 30 |
> | 25 | 3 | 75 |
> | 35 | 8 | 280 |
> | 45 | 14 | 630 |
> | 55 | 8 | 440 |
> | 65 | 3 | 195 |
> | 75 | 2 | 150 |
>
> $N = 2+3+8+14+8+3+2 = 40$. $\sum f_ix_i = 30+75+280+630+440+195+150 = 1800$.
>
> **Step 3 — Mean:**
> $$\bar x = \frac{\sum f_ix_i}{N} = \frac{1800}{40} = 45$$
>
> **Step 4 — Absolute deviations $|x_i-\bar x|$ and $f_i|x_i-\bar x|$:**
>
> | $x_i$ | $|x_i-45|$ | $f_i$ | $f_i|x_i-45|$ |
> |---|---|---|---|
> | 15 | 30 | 2 | 60 | …

Q.Find the mean deviation about the mean for the following data: Marks obtained: $10\text{-}20$ , $20\text{-}30$ , $30\text{-}40$ , $40\text{-}50$ , $50\text{-}60$ , $60\text{-}70$ , $70\text{-}80$ ; Number of Students: $2$ , $3$ , $8$ , $14$ , $8$ , $3$ , $2$ .

Concept understanding — Mean Deviation About Mean

Mean Deviation About Mean – The Intuition First

The Precise Definition

Step-by-Step Calculation

Why Use Absolute Values?

When Is This Used?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days