Match the LIST-I with LIST-II LIST-I (Population Mean (\(\mu\)) and…
2024
Match the LIST-I with LIST-II
LIST-I (Population Mean (\(\mu\)) and \(\frac{1}{N}\sum x^2\)) | LIST-II (Population Standard Deviation (\(\sigma\))) |
|---|---|
A. \(\mu\) = 3, \(\frac{1}{N}\sum x^2\) = 90 | I. 12 |
B. \(\mu\) = 4, \(\frac{1}{N}\sum x^2\) = 116 | II. 11 |
C. \(\mu\) = 5, \(\frac{1}{N}\sum x^2\) = 146 | III. 9 |
D. \(\mu\) = 6, \(\frac{1}{N}\sum x^2\) = 180 | IV. 10 |
Choose the correct answer from the options given below:
Answer: C. A-III, B-IV, C-II, D-I — Concept: For a population of N values the variance is the mean of the squares minus the square of the mean, \(\sigma^2 = \frac{1}{N}\sum x^2 - \mu^2\), and…
- A.
A-I, B-II, C-III, D-IV
- B.
A-II, B-III, C-IV, D-I
- C.
A-III, B-IV, C-II, D-I
- D.
A-IV, B-III, C-II, D-I
Show answer & explanation
Correct answer: C
Concept: For a population of N values the variance is the mean of the squares minus the square of the mean, \(\sigma^2 = \frac{1}{N}\sum x^2 - \mu^2\), and the standard deviation is its non-negative square root, \(\sigma = \sqrt{\sigma^2}\). The identity matters because it recovers the spread of a population from just two summary numbers, the mean and the mean of the squares, without ever seeing an individual data value.
Application: apply the identity to each LIST-I entry in turn.
For \(\mu = 3\) and \(\frac{1}{N}\sum x^2 = 90\): \(\sigma^2 = 90 - 3^2 = 90 - 9 = 81\), so \(\sigma = \sqrt{81} = 9\).
For \(\mu = 4\) and \(\frac{1}{N}\sum x^2 = 116\): \(\sigma^2 = 116 - 4^2 = 116 - 16 = 100\), so \(\sigma = \sqrt{100} = 10\).
For \(\mu = 5\) and \(\frac{1}{N}\sum x^2 = 146\): \(\sigma^2 = 146 - 5^2 = 146 - 25 = 121\), so \(\sigma = \sqrt{121} = 11\).
For \(\mu = 6\) and \(\frac{1}{N}\sum x^2 = 180\): \(\sigma^2 = 180 - 6^2 = 180 - 36 = 144\), so \(\sigma = \sqrt{144} = 12\).
Cross-check: rearranging the same identity gives \(\frac{1}{N}\sum x^2 = \sigma^2 + \mu^2\), so every standard deviation found above must rebuild the mean of squares printed beside its LIST-I entry.
LIST-I entry | \(\mu\) | \(\sigma\) | \(\sigma^2 + \mu^2\) | LIST-II |
|---|---|---|---|---|
A | 3 | 9 | 81 + 9 = 90 | III |
B | 4 | 10 | 100 + 16 = 116 | IV |
C | 5 | 11 | 121 + 25 = 146 | II |
D | 6 | 12 | 144 + 36 = 180 | I |
Each rebuilt value reproduces the mean of squares given for that entry, which confirms the four standard deviations. The matching is therefore A-III, B-IV, C-II, D-I.