Question

What ia the number of trials of a binomial distribution having mean and SD as 3 and 1.5 respectively

Answer 1

Answer:

12

Step-by-step explanation:

mean =np =3

s d =√npq=1.5

variance =npq =[1.5]^2=2.25

                 [np]q =2.25

            3q =2.25

               q =2.25/3 =2.25x4 /3x4=9/12 =3/4

              p = 1 -q = 1 - 3/4 =1/4

      but np= 3

            n [1/4] = 3

          n = 3x4 =12

Answer 2

The number of trials of a binomial distribution having mean and SD as 3 and 1.5 respectively is 12

Solution:

Given that,

$Mean = 3\\\\Standard\ deviation = 1.5$

We know that,

$Mean = n \times p\\\\SD = \sqrt{npq}$

Where,

n is the number of trials

p is the probability of a success on one trial

q denotes the probability of a failure on one trial

Therefore,

$\sqrt{npq} = 1.5\\\\\sqrt{3q} = 1.5\\\\3q = 2.25\\\\q = 0.75\\\\We\ know\ that\\\\p = 1 - q\\\\p = 1 - 0.75\\\\p = 0.25$

Find number of trials

$np = 3\\\\n \times 0.25 = 3\\\\n = 12$

Thus number of trials of a binomial distribution is 12

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