Find the mean, variance and standard deviation using short cut method:

| Classes | Frequency |
|---|---|
| 30-40 | 3 |
| 40-50 | 7 |
| 50-60 | 12 |
| 60-70 | 15 |
| 70-80 | 8 |
| 80-90 | 3 |
| 90-100 | 2 |

Question

Find the mean, variance and standard deviation using short cut method:

| Classes | Frequency |
|---|---|
| 30-40 | 3 |
| 40-50 | 7 |
| 50-60 | 12 |
| 60-70 | 15 |
| 70-80 | 8 |
| 80-90 | 3 |
| 90-100 | 2 |

Accepted Answer

>

> [!TLDR]
> Mean $=62$, Variance $=201$, Standard Deviation $=\sqrt{201}\approx14.18$.
>
> **Step 1 — Set up the shortcut-method table.**
> Take assumed mean $A=65$ (the midpoint of the modal class $60$–$70$) and class width $h=10$. For each class, the midpoint is $x_i$ and $d_i=\dfrac{x_i-A}{h}$.
>
> | Class | $f_i$ | $x_i$ | $d_i=\dfrac{x_i-65}{10}$ | $f_id_i$ | $f_id_i^2$ |
> |---|---|---|---|---|---|
> | 30–40 | 3 | 35 | $-3$ | $-9$ | $27$ |
> | 40–50 | 7 | 45 | $-2$ | $-14$ | $28$ |
> | 50–60 | 12 | 55 | $-1$ | $-12$ | $12$ |
> | 60–70 | 15 | 65 | $0$ | $0$ | $0$ |
> | 70–80 | 8 | 75 | $1$ | $8$ | $8$ |
> | 80–90 | 3 | 85 | $2$ | $6$ | $12$ |
> | 90–100 | 2 | 95 | $3$ | $6$ | $18$ |
>
> $$N=\sum f_i=3+7+12+15+8+3+2=50.$$
> $$\sum f_id_i=-9-14-12+0+8+6+6=-15.$$
> $$\sum f_id_i^2=27+28+12+0+8+12+18=105.$$
>
> **Step 2 — Mean.** …

Concept understanding — Variance and Standard Deviation of Data

Variance and Standard Deviation of Data

Class	$f_i$	$x_i$	$d_i=\dfrac{x_i-65}{10}$	$f_id_i$	$f_id_i^2$
30–40	3	35	$-3$	$-9$	$27$
40–50	7	45	$-2$	$-14$	$28$
50–60	12	55	$-1$	$-12$	$12$
60–70	15	65	$0$	$0$	$0$
70–80	8	75	$1$	$8$	$8$
80–90	3	85	$2$	$6$	$12$
90–100	2	95	$3$	$6$	$18$

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