Match List - I with List - II:

List - I [Population mean (μ) and (1/N)Σxi2]	List - II [Population standard deviation (σ)]
A. μ = 9, (1/N)Σxi2 = 117	I. 10
B. μ = 10, (1/N)Σxi2 = 200	II. 4
C. μ = 11, (1/N)Σxi2 = 202	III. 6
D. μ = 12, (1/N)Σxi2 = 160	IV. 9

Choose the correct answer from the options given below:सूची - I को सूची - II से सुमेलित कीजिए:

सूची - I [समष्टि माध्य (μ) तथा (1/N)Σxi2]	सूची - II [समष्टि मानक विचलन (σ)]
A. μ = 9, (1/N)Σxi2 = 117	I. 10
B. μ = 10, (1/N)Σxi2 = 200	II. 4
C. μ = 11, (1/N)Σxi2 = 202	III. 6
D. μ = 12, (1/N)Σxi2 = 160	IV. 9

नीचे दिए गए विकल्पों में से सही उत्तर चुनिए:

1. AA-III, B-IV, C-I, D-II
2. BA-II, B-I, C-IV, D-III
3. CA-III, B-I, C-IV, D-II
4. DA-I, B-IV, C-II, D-III

Pick an option to check yourself, or open the answer below.

Show answer & explanation

Correct answer: C. A-III, B-I, C-IV, D-II

Explanation

σ² = (1/N)Σx_i² − μ². A: 117 − 81 = 36, σ = 6 (III). B: 200 − 100 = 100, σ = 10 (I). C: 202 − 121 = 81, σ = 9 (IV). D: 160 − 144 = 16, σ = 4 (II). Hence A-III, B-I, C-IV, D-II.σ² = (1/N)Σx_i² − μ²। A: 117 − 81 = 36, σ = 6 (III)। B: 200 − 100 = 100, σ = 10 (I)। C: 202 − 121 = 81, σ = 9 (IV)। D: 160 − 144 = 16, σ = 4 (II)। अतः A-III, B-I, C-IV, D-II।