Question

Find the standard deviation, σ, for the binomial distribution which has the stated values of n and p. round your answer to the nearest hundredth. n = 1680; p = 0.57

Answer 1

σ ² = npq = np(1-p) = 1680*(0.57)*(1-0.57) = 411.768
=>
σ = sqrt(411.768) = 20.292 (to the nearest thousandth).

Answer 2

The standard deviation, $\sigma$ for the binomial distribution will be $20.292$.

What is binomial distribution ?

Binomial distribution is the probability of a particular outcome in a series when the outcome has two distinct possibilities, success or failure.

Variance $= \sigma^2 = npq$  

Standard deviation $(\sigma)=\sqrt{\sigma^2}$

Where, $n=$ number of trials,

$p=$ Success probability

$q=$ failure probability

We have,

$n = 1680$

$p = 0.57$

So,

$q=1-p$

$q=1-0.57=0.43$

So,

Variance $(\sigma^2 )= npq$

                       $=1680*0.57*0.43$  

Variance $(\sigma^2 )=411.768$

And,

Standard deviation $(\sigma)=\sqrt{\sigma^2}$

So,

$(\sigma)=\sqrt{411.768}$

$(\sigma)=20.292$

So, Standard deviation is $20.292$.

Hence, we can say that the standard deviation, $\sigma$ for the binomial distribution will be $20.292$.

To know more about binomial distribution click here

https://brainly.com/question/14565246

#SPJ2