Question

the number of hours students spent studying for a test are 2,4,4,5,7. calculate the sample standard deviation and interpret the results. a. 1.50, showing low variability in study times b. 1.41, showing high variability in study times c. 1.67, showing very low variability in study times d. 1.82, showing moderate variability in study times

Explanation:

Step1: Calculate the mean

The mean $\bar{x}$ is calculated as the sum of the data points divided by the number of data points. The data set is $2, 4, 4, 5, 7$. The sum is $2 + 4 + 4 + 5 + 7 = 22$. There are $n = 5$ data points. So, $\bar{x}=\frac{22}{5}=4.4$.

Step2: Calculate the squared differences from the mean

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

• For $x_1 = 2$: $(2 - 4.4)^2 = (-2.4)^2 = 5.76$
• For $x_2 = 4$: $(4 - 4.4)^2 = (-0.4)^2 = 0.16$
• For $x_3 = 4$: $(4 - 4.4)^2 = (-0.4)^2 = 0.16$
• For $x_4 = 5$: $(5 - 4.4)^2 = (0.6)^2 = 0.36$
• For $x_5 = 7$: $(7 - 4.4)^2 = (2.6)^2 = 6.76$

Step3: Calculate the sum of squared differences

Sum these squared differences: $5.76 + 0.16 + 0.16 + 0.36 + 6.76 = 13.2$.

Step4: Calculate the sample variance

The sample variance $s^2$ is given by $\frac{\sum (x_i - \bar{x})^2}{n - 1}$. Here, $n = 5$, so $n - 1 = 4$. Thus, $s^2=\frac{13.2}{4}=3.3$.

Step5: Calculate the sample standard deviation

The sample standard deviation $s$ is the square root of the sample variance. So, $s = \sqrt{3.3}\approx1.82$.

Step6: Interpret the result

A standard deviation of approximately $1.82$ indicates moderate variability in the study times because it's neither extremely small (which would indicate low variability) nor extremely large (which would indicate high variability).

Answer:

d. 1.82, showing moderate variability in study times