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B
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Difficulty:
45%
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Question Stats:
67% (01:16) correct
33%
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| Score | Frequency |
| k-4 | 1 |
| k-2 | 4 |
| k | 8 |
| k+2 | n |
The table above shows a frequency chart for the scores on a certain accounting quiz, where k and n are positive integers. What is the standard deviation of the scores on the accounting quiz?
(1) k = 5
(2) n =12
Score Frequency
k-4 1
k-2 4
k 8
k+2 n
k-4 1
k-2 4
k 8
k+2 n
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12 Nov 2024, 02:13
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XimeSol
An important property to note is that if we add or subtract a constant to each term in a set, the standard deviation does not change.
Thus, knowing the value of n alone would be enough to determine the standard deviation of the set. The value of k is irrelevant because it simply shifts each score by k, keeping the distribution shape unchanged.
This means we can analyze the scores as if they were:
| Score | Frequency |
| -4 | 1 |
| -2 | 4 |
| 0 | 8 |
| 2 | n |
Since the standard deviation remains the same regardless of k, we only need the value of n to calculate it. Therefore, the answer is B.
20 Nov 2023, 05:36
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XimeSol
Standard deviation denotes how much individual values in a data set differ from the mean (average) of the set. Put otherwise, standard deviation is a measure of the amount of variation or dispersion in a set of values.
Statement 1
1) k=5
This statement provides the value of \(k\), however, the information is not relevant as we do not know the frequency i.e. how many terms are present with the given value. We therefore cannot obtain the mean and find the standard deviation. Hence, this statement is not sufficient. We can eliminate A and D.
Statement 2
2) n =12
| Score | Frequency |
| k-4 | 1 |
| k-2 | 4 |
| k | 8 |
| k+2 | 12 |
The mean of the quiz = \(\frac{k - 4 + 4k - 8 + 8k + 12k + 24}{25} = \frac{25k + 12}{25} = k + \frac{12}{25}\)
We can find the distance each of the score elements is from the mean as each of the scores is given in terms of k. Hence, knowing the number of terms (frequency) of each element we can determine the standard deviation of the quiz.
This statement is sufficient.
Option B
