Question

The number of fish caught each day by a fisherman for 10 consecutive days was reported to be 62, 58, 56, 57, 62, 59, 62, 57, 64, 63. What is the standard deviation of this dataset?

• A. 2.9
• B. 8.44
• C. 3.60
• D. 76

Hint:

In competitive exams, be careful whether the question implies a "sample" or a "population". Usually, a small dataset like this is treated as a sample, so you divide the sum of squares by \(N-1\). If you divided by \(N\) (for a population), you would get \(\sqrt{7.64} \approx 2.76\), which is also close but 2.9 is a better fit for the sample calculation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

Step 2: Key Formula or Approach:
The formula for the sample standard deviation (\(s\)) is: \[ s = \sqrt{\frac{\sum_{i=1}^{N}(x_i - \bar{x})^2}{N-1}} \] where \(x_i\) are the individual data points, \(\bar{x}\) is the mean, and \(N\) is the number of observations.

Step 3: Detailed Calculation:
1. Data: 62, 58, 56, 57, 62, 59, 62, 57, 64, 63. \(N = 10\).
2. Calculate the mean (\(\bar{x}\)):
\[ \bar{x} = \frac{62+58+56+57+62+59+62+57+64+63}{10} = \frac{598}{10} = 59.8 \] 3. Calculate the sum of squared deviations \(\sum(x_i - \bar{x})^2\):
\((62-59.8)^2 = (2.2)^2 = 4.84\)
\((58-59.8)^2 = (-1.8)^2 = 3.24\)
\((56-59.8)^2 = (-3.8)^2 = 14.44\)
\((57-59.8)^2 = (-2.8)^2 = 7.84\)
\((62-59.8)^2 = (2.2)^2 = 4.84\)
\((59-59.8)^2 = (-0.8)^2 = 0.64\)
\((62-59.8)^2 = (2.2)^2 = 4.84\)
\((57-59.8)^2 = (-2.8)^2 = 7.84\)
\((64-59.8)^2 = (4.2)^2 = 17.64\)
\((63-59.8)^2 = (3.2)^2 = 10.24\)
Sum of squares = \(4.84+3.24+14.44+7.84+4.84+0.64+4.84+7.84+17.64+10.24 = 76.4\)
4. Calculate the sample variance (\(s^2\)):
\[ s^2 = \frac{76.4}{10-1} = \frac{76.4}{9} \approx 8.489 \] 5. Calculate the sample standard deviation (\(s\)):
\[ s = \sqrt{8.489} \approx 2.91 \]
Step 4: Final Answer:
The calculated standard deviation is approximately 2.91. This matches option (A).