Question

A candy company claims that 10% of its plain candies are orange, and a sample of 100 such candies is randomly selected,a. Find the mean and standard deviation for the number of orange candies in such groups of 100

Answer 1

It is important to note the type of distribution of such sampling.

To choose a candy, it can either be 'orange' or 'not orange'.

The sample size of 100 is large enough. Hence, the sampling describes Binomial Distribution.

The mean and standard deviation of a Binomial Distribution, respectively is given as:

$\begin{gathered} \mu=np \\ \sigma=\sqrt[]{np(1-p)}_{} \end{gathered}$

Where,

0. n= number of trials or samples

,

1. p=probability of success (orange candies)

As given in the sample, n=100, p=10% or 0.1

Substitute into the formula given:

$\mu=100\times0.1=10$$\sigma=\sqrt[]{100\times0.1(1-0.1)}=\sqrt[]{100\times0.1\times0.9}=\sqrt[]{9}=3$

Hence, the mean is 10 and the standard deviation is 3.