Question

find the standard deviation, $sigma$, for the binomial distribution which has the stated values of $n$ and $p$. round your answer to the nearest hundredth.
$n = 503; p = 0.7$

a. $sigma=13.55$
b. $sigma = 7.87$
c. $sigma=10.28$
d. $sigma = 14.40$

Explanation:

Step1: Recall binomial std - dev formula

The formula for the standard deviation $\sigma$ of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$.

Step2: Substitute given values

Given $n = 503$ and $p=0.7$, then $1 - p=1 - 0.7 = 0.3$. Substitute into the formula: $\sigma=\sqrt{503\times0.7\times0.3}$.

Step3: Calculate the value

First, calculate $503\times0.7\times0.3=503\times0.21 = 105.63$. Then, $\sigma=\sqrt{105.63}\approx10.28$.

Answer:

C. $\sigma = 10.28$