Question

for the following set of data, find the sample standard deviation, to the nearest 20, 32, 33, 33, 29, 29, 37, 27, 34

Explanation:

Step1: Calculate the mean

First, find the sum of the data set. The data is \(20, 32, 33, 33, 29, 29, 37, 27, 34\). The sum \( \sum x = 20 + 32 + 33 + 33 + 29 + 29 + 37 + 27 + 34\). Let's calculate that: \(20+32 = 52\), \(52+33 = 85\), \(85+33 = 118\), \(118+29 = 147\), \(147+29 = 176\), \(176+37 = 213\), \(213+27 = 240\), \(240+34 = 274\). There are \(n = 9\) data points. So the mean \( \bar{x}=\frac{\sum x}{n}=\frac{274}{9}\approx30.4444\).

Step2: Calculate the squared differences

For each data point \(x_i\), calculate \((x_i - \bar{x})^2\):
- For \(20\): \((20 - 30.4444)^2=(-10.4444)^2\approx109.085\)
- For \(32\): \((32 - 30.4444)^2=(1.5556)^2\approx2.420\)
- For \(33\): \((33 - 30.4444)^2=(2.5556)^2\approx6.531\) (there are two 33s, so we'll have two of these)
- For \(29\): \((29 - 30.4444)^2=(-1.4444)^2\approx2.086\) (there are two 29s, so two of these)
- For \(37\): \((37 - 30.4444)^2=(6.5556)^2\approx43.000\)
- For \(27\): \((27 - 30.4444)^2=(-3.4444)^2\approx11.864\)
- For \(34\): \((34 - 30.4444)^2=(3.5556)^2\approx12.642\)

Step3: Sum the squared differences

Now sum all these squared differences. Let's list them: \(109.085 + 2.420 + 6.531 + 6.531 + 2.086 + 2.086 + 43.000 + 11.864 + 12.642\). Calculating step by step: \(109.085+2.420 = 111.505\), \(111.505+6.531 = 118.036\), \(118.036+6.531 = 124.567\), \(124.567+2.086 = 126.653\), \(126.653+2.086 = 128.739\), \(128.739+43.000 = 171.739\), \(171.739+11.864 = 183.603\), \(183.603+12.642 = 196.245\). So \( \sum (x_i - \bar{x})^2\approx196.245\).

Step4: Calculate the sample variance

The sample variance \(s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1}\). Here, \(n - 1 = 8\), so \(s^2=\frac{196.245}{8}\approx24.5306\).

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^2}=\sqrt{24.5306}\approx4.95\) (to the nearest hundredth, or as needed. Wait, let's check the calculations more accurately. Wait, maybe I made a mistake in the sum of squared differences. Let's recalculate the sum of \(x\) again: \(20 + 32 = 52\), \(52+33 = 85\), \(85+33 = 118\), \(118+29 = 147\), \(147+29 = 176\), \(176+37 = 213\), \(213+27 = 240\), \(240+34 = 274\). That's correct. \(n = 9\), mean is \(274/9\approx30.4444\). Now let's recalculate each squared difference with more precision:

- \(20\): \((20 - 274/9)^2=( (180 - 274)/9 )^2=(-94/9)^2 = 8836/81\approx109.0864\)
- \(32\): \((32 - 274/9)^2=( (288 - 274)/9 )^2=(14/9)^2 = 196/81\approx2.420\)
- \(33\): \((33 - 274/9)^2=( (297 - 274)/9 )^2=(23/9)^2 = 529/81\approx6.5309\) (two of these: \(2\times529/81 = 1058/81\approx13.062\))
- \(29\): \((29 - 274/9)^2=( (261 - 274)/9 )^2=(-13/9)^2 = 169/81\approx2.0864\) (two of these: \(2\times169/81 = 338/81\approx4.1728\))
- \(37\): \((37 - 274/9)^2=( (333 - 274)/9 )^2=(59/9)^2 = 3481/81\approx42.9753\)
- \(27\): \((27 - 274/9)^2=( (243 - 274)/9 )^2=(-31/9)^2 = 961/81\approx11.8642\)
- \(34\): \((34 - 274/9)^2=( (306 - 274)/9 )^2=(32/9)^2 = 1024/81\approx12.6420\)

Now sum all these fractions:

\(8836/81 + 196/81 + 1058/81 + 338/81 + 3481/81 + 961/81 + 1024/81\)

First, sum the numerators: \(8836 + 196 + 1058 + 338 + 3481 + 961 + 1024\). Let's calculate:

\(8836+196 = 9032\)

\(9032+1058 = 10090\)

\(10090+338 = 10428\)

\(10428+3481 = 13909\)

\(13909+961 = 14870\)

\(14870+1024 = 15894\)

So the sum of squared differences is \(15894/81\approx196.2222\)

Now, the sample variance \(s^2=\frac{15894/81}{8}=\frac{15894}{81\times8}=\frac{15894}{648}\approx24.5278\)

Then the sample standard deviation \(s=\sqrt{24.5278}\approx4.95\) (to the nearest hundredth). If we round to the nearest tenth, it's 5.0, but let's check with a calculator approach. Alternatively, using the formula for sample standard deviation:

The formula for sample standard deviation is \(s=\sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\)

We can also use the shortcut formula: \(s=\sqrt{\frac{\sum x_i^2 - \frac{(\sum x_i)^2}{n}}{n - 1}}\)

Let's calculate \(\sum x_i^2\):

\(20^2 = 400\), \(32^2 = 1024\), \(33^2 = 1089\) (two of these: \(2\times1089 = 2178\)), \(29^2 = 841\) (two of these: \(2\times841 = 1682\)), \(37^2 = 1369\), \(27^2 = 729\), \(34^2 = 1156\)

Now sum these: \(400 + 1024 + 2178 + 1682 + 1369 + 729 + 1156\)

Calculating step by step: \(400+1024 = 1424\), \(1424+2178 = 3602\), \(3602+1682 = 5284\), \(5284+1369 = 6653\), \(6653+729 = 7382\), \(7382+1156 = 8538\)

Now, \(\frac{(\sum x_i)^2}{n}=\frac{274^2}{9}=\frac{75076}{9}\approx8341.7778\)

Then \(\sum x_i^2 - \frac{(\sum x_i)^2}{n}=8538 - 8341.7778 = 196.2222\)

Then \(s^2=\frac{196.2222}{8}\approx24.5278\), so \(s=\sqrt{24.5278}\approx4.95\) (to the nearest hundredth) or 5.0 to the nearest tenth.

Answer:

The sample standard deviation, to the nearest hundredth, is approximately \(4.95\) (or to the nearest tenth, \(5.0\)). If we follow the problem's instruction "to the nearest" (assuming nearest tenth or hundredth, but likely hundredth), the answer is approximately \(4.95\).