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 ($\mu$)

The data set is \(19, 21, 21, 21, 21, 23\). The number of data points \(n = 6\).
The sum of the data is \(19 + 21+21+21+21+23 = 116\).
The mean \(\mu=\frac{116}{6}=\frac{58}{3}\approx19.333\) (wait, no, wait: \(19 + 21\times4+23=19 + 84+23 = 126\)? Wait, let's recalculate: \(19+21 = 40\), \(40+21 = 61\), \(61+21 = 82\), \(82+21 = 103\), \(103+23 = 126\). Oh, I made a mistake earlier. So \(n = 6\), sum \(= 126\), so mean \(\mu=\frac{126}{6}=21\).

Step2: Calculate the squared differences from the mean

For each data point \(x\), calculate \((x - \mu)^2\):

• For \(x = 19\): \((19 - 21)^2=(-2)^2 = 4\)
• For \(x = 21\) (four times): \((21 - 21)^2=0^2 = 0\) (four times, so total \(4\times0 = 0\))
• For \(x = 23\): \((23 - 21)^2=(2)^2 = 4\)

Step3: Calculate the sum of squared differences

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

Step4: Calculate the variance ($\sigma^2$)

Variance for population (since it's a group of data items, assuming population) is \(\sigma^2=\frac{\sum(x - \mu)^2}{n}=\frac{8}{6}=\frac{4}{3}\approx1.333\)

Step5: Calculate the standard deviation ($\sigma$)

Standard deviation is the square root of variance: \(\sigma=\sqrt{\frac{4}{3}}\approx1.1547\)

Step6: Round to two decimal places

Rounding \(1.1547\) to two decimal places gives \(1.15\) (wait, wait: \(\sqrt{\frac{4}{3}}\approx1.1547\), so to two decimal places, look at the third digit: 4, which is less than 5, so we round down? Wait, no: \(1.1547\), the first decimal is 1, second is 5, third is 4. So rounding to two decimals: \(1.15\)? Wait, no, wait: \(\frac{8}{6}=\frac{4}{3}\approx1.333\)? Wait, no, wait: sum of squared differences is 8, n=6, so variance is 8/6 ≈1.333, then standard deviation is sqrt(1.333)≈1.1547, which is approximately 1.15 when rounded to two decimal places? Wait, no, 1.1547: the third decimal is 4, so we keep the second decimal as 5? Wait, 1.1547: first decimal 1, second 5, third 4. So rounding to two decimals: 1.15? Wait, no, 1.1547 is closer to 1.15 than 1.16? Wait, 1.1547 - 1.15 = 0.0047, 1.16 - 1.1547 = 0.0053. So since 0.0047 < 0.0053, we round to 1.15? Wait, but let's recalculate the sum of squared differences again. Wait, data points: 19,21,21,21,21,23. So mean is 21. So (19-21)^2=4, (21-21)^2=0 (four times), (23-21)^2=4. So sum is 4 + 04 +4=8. n=6. So variance is 8/6=1.333..., standard deviation is sqrt(8/6)=sqrt(4/3)≈1.1547. So to two decimal places, 1.15. Wait, but maybe I made a mistake in the mean. Wait, 19+21+21+21+21+23: 19+23=42, 214=84, total 42+84=126. 126/6=21. Correct. So sum of squared differences: (19-21)^2=4, (23-21)^2=4, and four (21-21)^2=0. So total 8. Variance 8/6=4/3≈1.333. Std dev sqrt(4/3)≈1.1547≈1.15.

Answer:

\(1.15\)