Question

is a binomial variate with parameter 15 and 1/3, what is the value of mode of thedistribution?(a) 5 and 6.(d) 6.(b) 5.(c) 5.50What is the number of trials of a binomial distribution having mean and SD as 3 and 1.5​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

QUESTION : 1

In is a binomial variate with parameter 15 and 1/3, what is the value of mode of the distribution

(a) 5 and 6

(b) 5

(c) 5.50

(d) 6

QUESTION : 2

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

CONCEPT TO BE IMPLEMENTED

Binomial Distribution

If a trial is repeated n times and p is the probability of a success and q that of failure then the probability of r successes is

$\displaystyle \sf{ \sf{P(X=r) = \: \: }\large{ {}^{n} C_r}\: {p}^{r} \: \: {q}^{n - r} } \: \: \: \: \: where \: q \: = 1 - p$

EVALUATION

ANSWER TO QUESTION : 1

Here it is given that in binomial variate with parameter 15 and 1/3

So n = 15 & p = 1/3

Thus ( n + 1 )p

$\displaystyle\sf{ = (15 + 1) \times \frac{1}{3} }$

$\displaystyle\sf{ = 16 \times \frac{1}{3} }$

$\displaystyle\sf{ =5.33}$

Since 5.33 is not an integer

Required mode = [ 5.33 ] = 5

Hence the correct option is (b) 5

ANSWER TO QUESTION : 2

Here it is given that in a binomial distribution having mean and SD as 3 and 1.5

Let the parameters are n and p

∴ Mean = np = 3 - - - - (1)

Again SD = 1.5

⇒ Variance = 1.5² = 2.25

⇒ npq = 2.25 - - - - (2)

Equation 2 ÷ Equation 1 gives

q = 0.75

⇒ p = 1 - q = 1 - 0.75 = 0.25

From Equation 1 we get

n × 0.25 = 3

⇒ n = 12

Hence the required number of trials = 12

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