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The standard deviation of 4, 5, 6, 7, .....13 is x, then the standard deviation of 14, 15, 16....23 is
- a)x
- b)10x
- c)x + 10
- d)x + √10
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The standard deviation of 4, 5, 6, 7, .....13 is x, then the standard ...
We can use the formula for standard deviation of a set of numbers:
$\text{SD} = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}}$
where $\bar{x}$ is the mean of the set and $n$ is the number of elements in the set.
For the first set of numbers (4, 5, 6, 7, ..., 13), we have $\bar{x} = 8.5$ and $n = 10$ (since there are 10 numbers in the set). We can calculate the sum of squares of deviations from the mean as:
$(4-8.5)^2 + (5-8.5)^2 + \cdots + (13-8.5)^2 = 82.5$
So the standard deviation of this set is:
$x = \sqrt{\frac{82.5}{10}} = \sqrt{8.25} \approx 2.872$
For the second set of numbers (14, 15, 16, ..., 23), we have $\bar{x} = 18.5$ and $n = 10$ again. Using the same formula, we can calculate the sum of squares of deviations from the mean as:
$(14-18.5)^2 + (15-18.5)^2 + \cdots + (23-18.5)^2 = 82.5$
Notice that the sum of squares of deviations for this set is the same as the first set! This is not a coincidence, since both sets have the same "shape", just shifted by 10. Therefore, the standard deviation of this set is also $x = \sqrt{8.25} \approx 2.872$.
So the answer is $\boxed{\text{(a) }x}$.
$\text{SD} = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}}$
where $\bar{x}$ is the mean of the set and $n$ is the number of elements in the set.
For the first set of numbers (4, 5, 6, 7, ..., 13), we have $\bar{x} = 8.5$ and $n = 10$ (since there are 10 numbers in the set). We can calculate the sum of squares of deviations from the mean as:
$(4-8.5)^2 + (5-8.5)^2 + \cdots + (13-8.5)^2 = 82.5$
So the standard deviation of this set is:
$x = \sqrt{\frac{82.5}{10}} = \sqrt{8.25} \approx 2.872$
For the second set of numbers (14, 15, 16, ..., 23), we have $\bar{x} = 18.5$ and $n = 10$ again. Using the same formula, we can calculate the sum of squares of deviations from the mean as:
$(14-18.5)^2 + (15-18.5)^2 + \cdots + (23-18.5)^2 = 82.5$
Notice that the sum of squares of deviations for this set is the same as the first set! This is not a coincidence, since both sets have the same "shape", just shifted by 10. Therefore, the standard deviation of this set is also $x = \sqrt{8.25} \approx 2.872$.
So the answer is $\boxed{\text{(a) }x}$.
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The standard deviation of 4, 5, 6, 7, .....13 is x, then the standard deviation of 14, 15, 16....23 isa)xb)10xc)x + 10d)x +√10Correct answer is option 'A'. Can you explain this answer?
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