Question

a local county has an unemployment rate of 5%. a random sample of 36 employable people are picked at random from the county and are asked if they are employed. round answers to 4 decimal places.
a) find the probability that exactly 5 in the sample are unemployed.
b) find the probability that there are fewer than 2 in the sample are unemployed.
c) find the probability that there are more than 4 in the sample are unemployed.
d) find the probability that there are at most 2 in the sample are unemployed.

Explanation:

Step1: Identify binomial distribution parameters

The binomial distribution formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 36$, $p=0.05$, and $1 - p = 0.95$.

Step2: Calculate probability for part a

For $k = 5$:
First, calculate the combination $C(36,5)=\frac{36!}{5!(36 - 5)!}=\frac{36!}{5!×31!}=\frac{36\times35\times34\times33\times32}{5\times4\times3\times2\times1}=376992$.
Then, $P(X = 5)=C(36,5)\times(0.05)^{5}\times(0.95)^{31}$
$P(X = 5)=376992\times(0.05)^{5}\times(0.95)^{31}\approx376992\times3.125\times10^{-7}\times0.2183=0.0256$.

Step3: Calculate probability for part b

$P(X\lt2)=P(X = 0)+P(X = 1)$
$C(36,0)=\frac{36!}{0!(36 - 0)!}=1$, $P(X = 0)=C(36,0)\times(0.05)^{0}\times(0.95)^{36}=1\times1\times0.1673 = 0.1673$
$C(36,1)=\frac{36!}{1!(36 - 1)!}=36$, $P(X = 1)=C(36,1)\times(0.05)^{1}\times(0.95)^{35}=36\times0.05\times0.1761=0.3170$
$P(X\lt2)=0.1673 + 0.3170=0.4843$.

Step4: Calculate probability for part c

$P(X\gt4)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$
$C(36,2)=\frac{36!}{2!(36 - 2)!}=\frac{36\times35}{2\times1}=630$, $P(X = 2)=C(36,2)\times(0.05)^{2}\times(0.95)^{34}=630\times0.0025\times0.1854 = 0.2907$
$C(36,3)=\frac{36!}{3!(36 - 3)!}=7140$, $P(X = 3)=C(36,3)\times(0.05)^{3}\times(0.95)^{33}=7140\times1.25\times10^{-4}\times0.1952=0.1736$
$C(36,4)=\frac{36!}{4!(36 - 4)!}=58905$, $P(X = 4)=C(36,4)\times(0.05)^{4}\times(0.95)^{32}=58905\times6.25\times10^{-6}\times0.2055=0.0754$
$P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)=0.1673+0.3170 + 0.2907+0.1736+0.0754=1 - 0.0760=0.9240$
$P(X\gt4)=1 - 0.9240=0.0760$.

Step5: Calculate probability for part d

$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)$
$P(X = 0)+P(X = 1)+P(X = 2)=0.1673+0.3170 + 0.2907=0.7750$.

Answer:

a) $0.0256$
b) $0.4843$
c) $0.0760$
d) $0.7750$