Question

bell ringer - oct 01

1. people who have some college education fifty - three percent of all persons in the u.s. population have at least some college education. choose 10 persons at random. find the probability that

a. exactly one - half have some college education
b. at least 6 do not have any college education
c. fewer than 4 have some college education

1. alternate sources of fuel: 85% percent of americans favor spending government money to develop alternative sources of fuel for automobiles. for a random sample of 140 americans, find the mean, variance, and standard deviation for the number who favor government spending for alternative fuels.

Explanation:

Step1: Identify binomial distribution parameters for part 1

Let $n = 10$ (number of trials/sample size). For people with college - education, $p=0.53$ (probability of success).

Step2: Solve part 1a

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$. For $k = 5$ (exactly one - half), $C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$. Then $P(X = 5)=252\times(0.53)^{5}\times(1 - 0.53)^{10 - 5}=252\times(0.53)^{5}\times(0.47)^{5}\approx252\times0.0416\times0.0229\approx0.242$.

Step3: Solve part 1b

The probability that a person does not have college education is $q = 1 - p=1 - 0.53 = 0.47$. We want $P(X\geq6)$ where $X$ is the number of people without college education. $P(X\geq6)=\sum_{k = 6}^{10}C(10,k)\times(0.47)^{k}\times(0.53)^{10 - k}=C(10,6)\times(0.47)^{6}\times(0.53)^{4}+C(10,7)\times(0.47)^{7}\times(0.53)^{3}+C(10,8)\times(0.47)^{8}\times(0.53)^{2}+C(10,9)\times(0.47)^{9}\times(0.53)^{1}+C(10,10)\times(0.47)^{10}\times(0.53)^{0}$.
$C(10,6)=\frac{10!}{6!(10 - 6)!}=210$, $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$, $C(10,8)=\frac{10!}{8!(10 - 8)!}=45$, $C(10,9)=\frac{10!}{9!(10 - 9)!}=10$, $C(10,10)=\frac{10!}{10!(10 - 10)!}=1$.
$P(X\geq6)\approx0.237$.

Step4: Solve part 1c

We want $P(X\lt4)$ for people with college education. $P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)$.
$P(X = 0)=C(10,0)\times(0.53)^{0}\times(0.47)^{10}\approx0.0006$.
$C(10,1)=\frac{10!}{1!(10 - 1)!}=10$, $P(X = 1)=10\times(0.53)^{1}\times(0.47)^{9}\approx0.0064$.
$C(10,2)=\frac{10!}{2!(10 - 2)!}=45$, $P(X = 2)=45\times(0.53)^{2}\times(0.47)^{8}\approx0.0337$.
$C(10,3)=\frac{10!}{3!(10 - 3)!}=120$, $P(X = 3)=120\times(0.53)^{3}\times(0.47)^{7}\approx0.102$.
$P(X\lt4)\approx0.0006 + 0.0064+0.0337 + 0.102=0.1427$.

Step5: For part 2

Let $n = 140$ (sample size) and $p = 0.85$ (probability of favoring government spending for alternative fuels).
The mean of a binomial distribution is $\mu=np$. So $\mu=140\times0.85 = 119$.
The variance is $\sigma^{2}=np(1 - p)=140\times0.85\times(1 - 0.85)=140\times0.85\times0.15 = 17.85$.
The standard deviation is $\sigma=\sqrt{np(1 - p)}=\sqrt{17.85}\approx4.22$.

Answer:

1a. Approximately $0.242$
1b. Approximately $0.237$
1c. Approximately $0.1427$

1. Mean: $119$, Variance: $17.85$, Standard Deviation: Approximately $4.22$