Question

how do you calculate the standard deviation and the population deviation for the following You will take a sample of 27 of them for different weeks and you observe the following numbers of hours that they work during a week.numbers? 73,19,16,64,28,28,31,90,60,56,31,56,22,18,45,48,17,17,17,91,92,63,50,51,69,16,17

Answer 1

To find standard deviation we need to find sample mean first

$\overline{x}=\frac{\sum x}{n}=43.8889$

Create the following table.

data	data-mean	(data - mean)2
73	29.1111	847.45614321
19	-24.8889	619.45734321
16	-27.8889	777.79074321
64	20.1111	404.45634321
28	-15.8889	252.45714321
28	-15.8889	252.45714321
31	-12.8889	166.12374321
90	46.1111	2126.23354321
60	16.1111	259.56754321
56	12.1111	146.67874321
31	-12.8889	166.12374321
56	12.1111	146.67874321
22	-21.8889	479.12394321
18	-25.8889	670.23514321
45	1.1111	1.23454321
48	4.1111	16.90114321
17	-26.8889	723.01294321
17	-26.8889	723.01294321
17	-26.8889	723.01294321
91	47.1111	2219.45574321
92	48.1111	2314.67794321
63	19.1111	365.23414321
50	6.1111	37.34554321
51	7.1111	50.56774321
69	25.1111	630.56734321
16	-27.8889	777.79074321
17	-26.8889	723.01294321

Find the sum of numbers in the last column to get. $\sum (xi-\overline{x})^2=16620.6667$

So Population standard deviation in $\sigma=\sqrt{\frac{\sum (xi-\overline{x})^2}{n}}=\sqrt{\frac{16620.6667}{27}}=24.8109$

And sample standard deviation is $s=\sqrt{\frac{\sum (xi-\overline{x})^2}{n-1}}=\sqrt{\frac{16620.6667}{27-1}}=25.2835$