Question

Calculate the standard deviation of the given temperature below recorded for the 7 days of a week:
24, 24, 21, 20, 26, 27, 30

• A. 24
• B. 3.2
• C. 24.57
• D. 11.95

Hint:

Notice the options:
24.57 is the mean.
11.95 is the sample variance \(\frac{71.71}{6}\).
3.2 is the standard deviation (\(\sqrt{10.24} = 3.2\)).

Answer 1

The Correct Option is B

Concept:
Standard deviation (\(\sigma\)) is a statistical measure of the dispersion or variation of a set of data values relative to their mean.
It is computed as the square root of the variance.

Step 1: Key Formula:
The mean (\(\mu\)) of \(N\) values is: \[ \mu = \frac{\sum x_i}{N} \] The population standard deviation (\(\sigma\)) is given by: \[ \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} \]

Step 2: Step-by-Step Computation:
The given data values for \(N = 7\) days are: \[ 24, 24, 21, 20, 26, 27, 30 \] Calculate the sum of values: \[ \sum x_i = 24 + 24 + 21 + 20 + 26 + 27 + 30 = 172 \] Calculate the arithmetic mean: \[ \mu = \frac{172}{7} \approx 24.57 \] Compute the deviations and their squares: \[ (24 - 24.57)^2 = (-0.57)^2 \approx 0.325 \] \[ (24 - 24.57)^2 = (-0.57)^2 \approx 0.325 \] \[ (21 - 24.57)^2 = (-3.57)^2 \approx 12.745 \] \[ (20 - 24.57)^2 = (-4.57)^2 \approx 20.885 \] \[ (26 - 24.57)^2 = (1.43)^2 \approx 2.045 \] \[ (27 - 24.57)^2 = (2.43)^2 \approx 5.905 \] \[ (30 - 24.57)^2 = (5.43)^2 \approx 29.485 \] Sum of squared deviations: \[ \sum (x_i - \mu)^2 \approx 0.325 + 0.325 + 12.745 + 20.885 + 2.045 + 5.905 + 29.485 = 71.71 \] Calculate the variance: \[ \sigma^2 = \frac{71.71}{7} \approx 10.244 \] Calculate the standard deviation: \[ \sigma = \sqrt{10.244} \approx 3.20 \] Final Answer:
The standard deviation of the recorded temperatures is approximately 3.2. Thus, option (B) is the correct answer.