Match the LIST-I with LIST-II

LIST-I (n = number of Trials, p = Probability of Success)	LIST-II (Mean and Standard Deviation of Binomial Distribution)
A. n = 5, p = 0.4	I. 2.8 and 1.296
B. n = 6, p = 0.6	II. 2.0 and 1.095
C. n = 8, p = 0.5	III. 4.0 and 1.414
D. n = 7, p = 0.4	IV. 3.6 and 1.200

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

सूची-I (n = परीक्षणों की संख्या, p = सफलता की प्रायिकता)	सूची-II (द्विपद बंटन का माध्य एवं मानक विचलन)
A. n = 5, p = 0.4	I. 2.8 और 1.296
B. n = 6, p = 0.6	II. 2.0 और 1.095
C. n = 8, p = 0.5	III. 4.0 और 1.414
D. n = 7, p = 0.4	IV. 3.6 और 1.200

नीचे दिए गए विकल्पों में से सही उत्तर चुनिए:

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

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

Show answer & explanation

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

Explanation

For a binomial distribution, mean = np and SD = √(npq). A: mean = 2.0, SD = √1.2 = 1.095 (II). B: mean = 3.6, SD = √1.44 = 1.200 (IV). C: mean = 4.0, SD = √2 = 1.414 (III). D: mean = 2.8, SD = √1.68 = 1.296 (I). Hence A-II, B-IV, C-III, D-I.द्विपद बंटन में माध्य = np तथा मानक विचलन = √(npq)। A: माध्य = 2.0, मा.वि. = √1.2 = 1.095 (II)। B: माध्य = 3.6, मा.वि. = √1.44 = 1.200 (IV)। C: माध्य = 4.0, मा.वि. = √2 = 1.414 (III)। D: माध्य = 2.8, मा.वि. = √1.68 = 1.296 (I)। अतः A-II, B-IV, C-III, D-I।