Question

calculate the standard deviation and variance of the sample quantitative data shown, to two decimal places.
x
9.1
29.6
8.8
8.9
19.1
standard deviation:
variance:

Explanation:

Step1: Calculate the mean ($\bar{x}$)

The formula for the mean of a sample is $\bar{x} = \frac{\sum_{i = 1}^{n} x_i}{n}$, where $n$ is the number of data points and $x_i$ are the data values.
Here, $n = 5$, and the data points are $9.1, 29.6, 8.8, 8.9, 19.1$.
$\sum_{i = 1}^{5} x_i = 9.1 + 29.6 + 8.8 + 8.9 + 19.1 = 75.5$
$\bar{x} = \frac{75.5}{5} = 15.1$

Step2: Calculate the squared differences from the mean

For each data point $x_i$, calculate $(x_i - \bar{x})^2$:

• For $x_1 = 9.1$: $(9.1 - 15.1)^2 = (-6)^2 = 36$
• For $x_2 = 29.6$: $(29.6 - 15.1)^2 = (14.5)^2 = 210.25$
• For $x_3 = 8.8$: $(8.8 - 15.1)^2 = (-6.3)^2 = 39.69$
• For $x_4 = 8.9$: $(8.9 - 15.1)^2 = (-6.2)^2 = 38.44$
• For $x_5 = 19.1$: $(19.1 - 15.1)^2 = (4)^2 = 16$

Step3: Calculate the sum of squared differences

$\sum_{i = 1}^{5} (x_i - \bar{x})^2 = 36 + 210.25 + 39.69 + 38.44 + 16 = 340.38$

Step4: Calculate the sample variance ($s^2$)

The formula for sample variance is $s^2 = \frac{\sum_{i = 1}^{n} (x_i - \bar{x})^2}{n - 1}$
Here, $n - 1 = 4$
$s^2 = \frac{340.38}{4} = 85.095 \approx 85.10$ (rounded to two decimal places)

Step5: Calculate the sample standard deviation ($s$)

The standard deviation is the square root of the variance, so $s = \sqrt{s^2}$
$s = \sqrt{85.095} \approx 9.22$ (rounded to two decimal places)

Answer:

Standard deviation: $9.22$
Variance: $85.10$