Question

find the standard deviation for the group of data items. 19, 21, 21, 21, 21, 23 the standard deviation is (simplify your answer. round to two decimal places as needed.)

Explanation:

Step1: Calculate the mean

The data set is \(19, 21, 21, 21, 21, 23\). The number of data points \(n = 6\).
The mean \(\bar{x}=\frac{19 + 21+21+21+21+23}{6}=\frac{136}{6}=\frac{68}{3}\approx22.67\)

Step2: Calculate the squared differences from the mean

For \(19\): \((19 - \frac{68}{3})^2=(\frac{57 - 68}{3})^2=(-\frac{11}{3})^2=\frac{121}{9}\)
For each \(21\) (there are 4 of them): \((21 - \frac{68}{3})^2=(\frac{63 - 68}{3})^2=(-\frac{5}{3})^2=\frac{25}{9}\)
For \(23\): \((23 - \frac{68}{3})^2=(\frac{69 - 68}{3})^2=(\frac{1}{3})^2=\frac{1}{9}\)

Step3: Calculate the sum of squared differences

Sum \(=\frac{121}{9}+4\times\frac{25}{9}+\frac{1}{9}=\frac{121 + 100+1}{9}=\frac{222}{9}=\frac{74}{3}\approx24.67\)

Step4: Calculate the variance

Variance \(s^2=\frac{\text{Sum of squared differences}}{n - 1}=\frac{\frac{74}{3}}{5}=\frac{74}{15}\approx4.93\)

Step5: Calculate the standard deviation

Standard deviation \(s=\sqrt{\frac{74}{15}}\approx\sqrt{4.93}\approx2.22\)

(Correct):

Step1: Calculate the mean

The data set is \(19, 21, 21, 21, 21, 23\). The number of data points \(n = 6\).
Mean \(\bar{x}=\frac{19 + 21+21+21+21+23}{6}=\frac{126}{6}=21\).

Step2: Find squared deviations from the mean

• For \(19\): \((19 - 21)^2=(-2)^2 = 4\)
• For each \(21\) (4 times): \((21 - 21)^2=0^2 = 0\)
• For \(23\): \((23 - 21)^2=2^2 = 4\)

Step3: Sum the squared deviations

Sum \(=4 + 0+0+0+0 + 4=8\).

Step4: Calculate population variance (if treating as population)

Population variance \(\sigma^2=\frac{\text{Sum of squared deviations}}{n}=\frac{8}{6}=\frac{4}{3}\approx1.33\).

Step5: Calculate population standard deviation

Standard deviation \(\sigma=\sqrt{\frac{4}{3}}\approx1.15\) (rounded to two decimal places).

Answer:

\(1.15\) (Wait, there is a mistake in the previous calculation. Let's recalculate:

Correct Step1: Mean \(\bar{x}=\frac{19 + 21+21+21+21+23}{6}=\frac{19+21\times4 + 23}{6}=\frac{19 + 84+23}{6}=\frac{126}{6} = 21\)

Correct Step2: Squared differences:
For \(19\): \((19 - 21)^2=(- 2)^2 = 4\)
For each \(21\) (4 times): \((21 - 21)^2=0\)
For \(23\): \((23 - 21)^2=2^2 = 4\)

Correct Step3: Sum of squared differences \(=4+0+0+0+0 + 4=8\)

Correct Step4: Variance \(s^2=\frac{8}{6 - 1}=\frac{8}{5}=1.6\)

Correct Step5: Standard deviation \(s=\sqrt{1.6}\approx1.26\)? Wait, no:

Wait, data set: \(19,21,21,21,21,23\)

Number of data points \(n = 6\)

Mean \(\bar{x}=\frac{19+21+21+21+21+23}{6}=\frac{19 + 21\times4+23}{6}=\frac{19 + 84 + 23}{6}=\frac{126}{6}=21\)

Squared deviations:

\((19 - 21)^2=(-2)^2 = 4\)

\((21 - 21)^2=0\) (four times)

\((23 - 21)^2=2^2 = 4\)

Sum of squared deviations \(=4 + 0+0+0+0 + 4=8\)

Variance (sample standard deviation, so divide by \(n - 1=5\)): \(s^2=\frac{8}{5}=1.6\)

Standard deviation \(s=\sqrt{1.6}\approx1.26\)? Wait, no, \(\sqrt{1.6}\approx1.2649\), rounded to two decimal places is \(1.26\)? Wait, no, let's check again:

Wait, the data is \(19,21,21,21,21,23\). Let's list the data:

Data points: \(x_1 = 19\), \(x_2=21\), \(x_3 = 21\), \(x_4=21\), \(x_5=21\), \(x_6 = 23\)

Mean \(\bar{x}=\frac{\sum_{i = 1}^{6}x_i}{6}=\frac{19 + 21+21+21+21+23}{6}=\frac{126}{6}=21\)

Deviations from mean: \(19 - 21=-2\), \(21 - 21 = 0\) (four times), \(23 - 21=2\)

Squared deviations: \((-2)^2 = 4\), \(0^2=0\) (four times), \(2^2 = 4\)

Sum of squared deviations: \(4+0+0+0+0 + 4=8\)

Sample variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{8}{5}=1.6\)

Sample standard deviation \(s=\sqrt{1.6}\approx1.26\)

Wait, but maybe it's population standard deviation? If it's population, then divide by \(n = 6\), variance \(=\frac{8}{6}=\frac{4}{3}\approx1.33\), standard deviation \(=\sqrt{\frac{4}{3}}\approx1.15\)

Ah, the problem says "standard deviation" without specifying sample or population. If it's population (since it's a group of data items, maybe population), then:

Population variance \(\sigma^2=\frac{8}{6}=\frac{4}{3}\)

Population standard deviation \(\sigma=\sqrt{\frac{4}{3}}\approx1.15\)

So the correct answer is \(1.15\))