# Suppose that the chance that a prospective employee will have a desirable quality is 0.3, and 16...

## Question:

Suppose that the chance that a prospective employee will have a desirable quality is 0.3, and 16 people apply for a particular job.

a) Does this satisfy the four requirements for a binomial problem?

b) Find the mean and standard deviation.

c) What is the chance that 5 of the individuals interviewed have this desired quality?

d) What is the chance that 3 or more of the individuals have this desired quality?

## Binomial Distribution:

Binomial distribution is a type of discrete probability distribution that gives sequence of successes repeated in a given number of Bernoulli trials. The probability of success remains constant throughout the experiment.

## Answer and Explanation:

**a).**

Yes. The requirements for binomial problem are satisfied. There are only two possible outcomes (prospective employee having a desirable quality and not having a desirable quality) and number of trials are more than 1 (16 peoples who applied the job).

**b).**

The mean and standard deviation of binomial distribution are calculated as;

{eq}\begin{align*} E(x)&=\mu=np\\&=0.3\times 16\\&=4.8\\\sigma&=\sqrt{np(1-p)}\\&=\sqrt{0.3\times 16(1-0.3)}\\&=1.833 \end{align*} {/eq}

**c).**

Use binomial probability mass function to calculate the following probability:

{eq}\begin{align*} P(X=5)&\\b(n,x,p)&=^nC_xp^x(1-p)^{n-x}\\b(16,5,0.3)&=^{16}C_5\cdot 0.3^5(1-0.3)^{16-5}\\&=4368\times 0.3^5\times 0.7^{11}\\&=0.2099 \end{align*} {/eq}

**d).**

{eq}\begin{align*} P(X\ge 3)&=1-\left[P(x=0)+P(x=1)+P(x=2)\right]\\P(x=0)&\\b(16,0,0.3)&=^{16}C_0\cdot 0.3^0(1-0.3)^{16-0}\\&=0.0033\\P(x=1)&\\b(16,1,0.3)&=^{16}C_1\cdot 0.3^1(1-0.3)^{16-1}\\&=0.0228\\P(x=2)&\\b(16,2,0.3)&=^{16}C_2\cdot 0.3^2(1-0.3)^{16-2}\\&=0.0732\\\therefore P(x\ge 3)&=1-(0.0033+0.0228+0.0732)\\&=0.9007 \end{align*} {/eq}

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