The number of customers served per hour at a bank counter, recorded over 30 hours, is given below. Find the Variance and Standard Deviation (a) by the direct method, and (b) by the step-deviation method (take $A=4$, $h=1$), and verify the two agree.

| Customers ($x$) | 2 | 3 | 4

Question

The number of customers served per hour at a bank counter, recorded over 30 hours, is given below. Find the Variance and Standard Deviation (a) by the direct method, and (b) by the step-deviation method (take $A=4$, $h=1$), and verify the two agree.

| Customers ($x$) | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Number of hours ($f$) | 4 | 7 | 10 | 6 | 3 |

Accepted Answer

**Step 1 — Direct method: find the mean and $\sum fx^2$.**

| $x$ | $f$ | $fx$ | $x^2$ | $fx^2$ |
|---|---|---|---|---|
| 2 | 4 | 8 | 4 | 16 |
| 3 | 7 | 21 | 9 | 63 |
| 4 | 10 | 40 | 16 | 160 |
| 5 | 6 | 30 | 25 | 150 |
| 6 | 3 | 18 | 36 | 108 |
| **Total** | **30** | **117** | | **497** |

$\bar{x} = 117/30 = 3.9$.
$$\sigma^2 = \dfrac{497}{30} - (3.9)^2 = 16.5667 - 15.21 = 1.3567, \qquad \sigma = \sqrt{1.3567} \approx 1.1648$$

**Step 2 — Cross-check by the step-deviation method, $A=4$, $h=1$.** Since $h=1$, $u=(x-A)/h$ is simply $x-4$.

| $x$ | $f$ | $u=x-4$ | $fu$ | $fu^2$ |
|---|---|---|---|---|
| 2 | 4 | $-2$ | $-8$ | 16 |
| 3 | 7 | $-1$ | $-7$ | 7 |
| 4 | 10 | 0 | 0 | 0 |
| 5 | 6 | 1 | 6 | 6 |
| 6 | 3 | 2 | 6 | 12 |
| **Total** | **30** | | **$-3$** | **41** | …

$x$	$f$	$fx$	$x^2$	$fx^2$
2	4	8	4	16
3	7	21	9	63
4	10	40	16	160
5	6	30	25	150
6	3	18	36	108
Total	30	117		497

$x$	$f$	$u=x-4$	$fu$	$fu^2$
2	4	$-2$	$-8$	16
3	7	$-1$	$-7$	7
4	10	0	0	0
5	6	1	6	6
6	3	2	6	12
Total	30		$-3$	41

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