Question

1. two data sets are given:

data set a: 1, 10, 20, 30, 40
data set b: 20, 22, 24, 26, 28
what are the sample standard deviations of data set a and data set b?

a. 14.36, 2.33

b. 14.36, 3.33

c. 15.50, 3.16

d. 13.36, 2.83

Explanation:

Step1: Recall the formula for sample standard deviation

The formula for sample standard deviation is $s = \sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $\bar{x}$ is the sample mean, $x_{i}$ are the data points, and $n$ is the number of data points.

Step2: Calculate the mean of Data Set A

Data Set A: \(1, 10, 20, 30, 40\)
The mean $\bar{x}_{A}=\frac{1 + 10+20 + 30+40}{5}=\frac{101}{5} = 20.2$

Step3: Calculate the sum of squared deviations for Data Set A

For each data point $x_{i}$ in Data Set A:
- For \(x = 1\): \((1 - 20.2)^{2}=(- 19.2)^{2}=368.64\)
- For \(x = 10\): \((10 - 20.2)^{2}=(-10.2)^{2}=104.04\)
- For \(x = 20\): \((20 - 20.2)^{2}=(-0.2)^{2}=0.04\)
- For \(x = 30\): \((30 - 20.2)^{2}=(9.8)^{2}=96.04\)
- For \(x = 40\): \((40 - 20.2)^{2}=(19.8)^{2}=392.04\)

Sum of squared deviations: \(368.64+104.04 + 0.04+96.04+392.04=960.8\)

Step4: Calculate the sample standard deviation for Data Set A

Using the formula \(s_{A}=\sqrt{\frac{960.8}{5 - 1}}=\sqrt{\frac{960.8}{4}}=\sqrt{240.2}\approx15.50\)? Wait, no, wait, I made a mistake in the mean calculation. Wait, \(1 + 10+20 + 30+40=101\)? Wait, \(1+10 = 11\), \(11 + 20=31\), \(31+30 = 61\), \(61 + 40=101\)? Wait, no, \(1+10+20+30+40=101\)? Wait, no, \(1+10 = 11\), \(11+20 = 31\), \(31 + 30=61\), \(61+40 = 101\). Wait, but \(101\div5 = 20.2\). But let's recalculate the sum of squared deviations:

Wait, maybe I made a mistake in the data points. Wait, Data Set A is \(1,10,20,30,40\). Let's recalculate the mean: \(\bar{x}_{A}=\frac{1 + 10+20+30+40}{5}=\frac{101}{5}=20.2\) (correct). Now, the squared deviations:

\((1 - 20.2)^2=( - 19.2)^2 = 368.64\)

\((10 - 20.2)^2=( - 10.2)^2=104.04\)

\((20 - 20.2)^2=( - 0.2)^2 = 0.04\)

\((30 - 20.2)^2=(9.8)^2 = 96.04\)

\((40 - 20.2)^2=(19.8)^2=392.04\)

Sum: \(368.64+104.04=472.68\); \(472.68 + 0.04 = 472.72\); \(472.72+96.04 = 568.76\); \(568.76+392.04 = 960.8\) (correct). Then \(s_{A}=\sqrt{\frac{960.8}{4}}=\sqrt{240.2}\approx15.50\)? Wait, but let's check Data Set B.

Data Set B: \(20,22,24,26,28\)

Step5: Calculate the mean of Data Set B

\(\bar{x}_{B}=\frac{20 + 22+24+26+28}{5}=\frac{120}{5}=24\)

Step6: Calculate the sum of squared deviations for Data Set B

For each data point:

- \(20 - 24=-4\), \((-4)^2 = 16\)
- \(22 - 24=-2\), \((-2)^2 = 4\)
- \(24 - 24 = 0\), \(0^2=0\)
- \(26 - 24 = 2\), \(2^2 = 4\)
- \(28 - 24 = 4\), \(4^2 = 16\)

Sum of squared deviations: \(16 + 4+0 + 4+16=40\)

Step7: Calculate the sample standard deviation for Data Set B

\(s_{B}=\sqrt{\frac{40}{5 - 1}}=\sqrt{\frac{40}{4}}=\sqrt{10}\approx3.16\)

Wait, but now looking at the options, option c is \(15.50,3.16\). But wait, earlier when I calculated Data Set A's standard deviation, I got approximately \(15.50\) and Data Set B's is approximately \(3.16\), which matches option c? Wait, but let's re - check Data Set A's mean. Wait, \(1+10+20+30+40 = 101\)? Wait, no, \(1+10 = 11\), \(11+20 = 31\), \(31+30 = 61\), \(61+40 = 101\). So mean is \(20.2\). But let's use another approach. The formula for sample standard deviation can also be calculated using the formula \(s=\sqrt{\frac{\sum x_{i}^{2}-\frac{(\sum x_{i})^{2}}{n}}{n - 1}}\)

For Data Set A:

\(\sum x_{i}=1 + 10+20+30+40 = 101\)

\(\sum x_{i}^{2}=1^{2}+10^{2}+20^{2}+30^{2}+40^{2}=1 + 100+400+900+1600 = 3001\)

Then \(\frac{(\sum x_{i})^{2}}{n}=\frac{101^{2}}{5}=\frac{10201}{5}=2040.2\)

\(\sum x_{i}^{2}-\frac{(\sum x_{i})^{2}}{n}=3001-2040.2 = 960.8\)

Then \(s_{A}=\sqrt{\frac{960.8}{4}}=\sqrt{240.2}\approx15.50\) (correct)

For Data Set B:

\(\sum x_{i}=20 + 22+24+26+28 = 120\)

\(\sum x_{i}^{2}=20^{2}+22^{2}+24^{2}+26^{2}+28^{2}=400 + 484+576+676+784 = 2920\)

\(\frac{(\sum x_{i})^{2}}{n}=\frac{120^{2}}{5}=\frac{14400}{5}=2880\)

\(\sum x_{i}^{2}-\frac{(\sum x_{i})^{2}}{n}=2920 - 2880=40\)

\(s_{B}=\sqrt{\frac{40}{4}}=\sqrt{10}\approx3.16\)

So the sample standard deviation of Data Set A is approximately \(15.50\) and Data Set B is approximately \(3.16\), which corresponds to option c.

Wait, but earlier when I thought I made a mistake, but after recalculating using the sum of squares formula, it's correct.

Answer:

c. 15.50, 3.16