Question

Determine the probability P(1 or fewer) for a binomial experiment with 2-12 trials and the success probability p=0.3. Then find the mean, variance, and standard deviation. B Part 1 of 3 !! 2 Determine the probability P(1 or fewer). Round the answer to at least four decimal places. P(1 or fewer)-D nh Part 2 of 3 Find the mean. If necessary, round the answer to two decimal places, The mean is Part 3 of 3 Find the variance and standard deviation. If necessary, round the variance to two decimal places and standard deviation to at least three decimal places. The variance is The standard deviation is

Answer 1

Answer:

Given that,

Determine the probability P(1 or fewer) for a binomial experiment with n=12 trials and the success probability p=0.3.

Then find the mean, variance, and standard deviation:

We know that,

$\rightarrow$ The binomial distribution formula,

P(X = 2) = nc, xp'(1 - p)"

Here,

n=12 p=0.3

(1).

P (1 or fewer):

P (1 or fewer) means P(X < 1).

P(X < 1)=P(X=0)

Then,

$P(X =0)=12_{C_0}\times (0.3)^{0}\times (1-0.3)^{12-0}$

=1x1x (0.7)12

=0.013842872

=0.0138 (Approximately)

Therefore, P( 1 or fewer)=0.0138.

(2).

The mean ($\mu _x$):

We know that the mean is $\mu _x=n \times p$

=12 $\times$ 0.3

=3.6

Therefore, the mean ($\mu _x$)=3.6.

(3).

The variance ($\sigma _x^2$):

$\rightarrow$ We know that the variance is,

$\sigma _x^2=n\times p\times (1-p)$

=12 $\times$ 0.3 $\times$ (1-0.3)

=12 $\times$ 0.3 $\times$ 0.7

=2.52

Therefore, the variance is ($\sigma _x^2$)=2.52.

The standard deviation ($\sigma _x$):

$\rightarrow$ We know that, the standard deviation is,

$\sigma _x=\sqrt{Variance}\ [or]\ \sqrt{n\times p\times (1-p)}$

$=\sqrt{2.52}$

=1.58745079

=1.588 (Approximately)

Therefore, the standard deviation is ($\sigma _x$)=1.588.