Answer to Question #151053 in Statistics and Probability for Nazatul Aisyah Abd Rahman

Question

Many U.S. households still do not have Internet access. Suppose 25 out of 80 households in a small southern town do not have Internet access. A company that provides high-speed Internet has recently entered the market. As part of the marketing campaign, the company decides to randomly select ten households and offer them red laptops along with a brochure that describes their services. The aim is to build goodwill and, with a free laptop, tempt nonusers into getting Internet access.
a. What is the probability that six laptop recipients do not have Internet access?
(2 Marks)
b. What is the probability that at least five recipients do not have Internet access?
(2 Marks)
c. What is the probability that two or fewer laptop recipients do not have Internet access?
(2 Marks)
d. What is the expected number of laptop recipients who do not have Internet access?
(1 Marks)
e. Calculate the expected value, the variance, and the standard deviation for the recipients who do not have Internet access.

Answer 1

Let's denote x for the number of households who do not have internet access,

n for the sample size, p for the probability of success.

Here

$n=80$

$x=25$

$p=\frac{25}{80}=0.3125$

Here x follows binomial distribution

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}$

a) n = 10

$P(X=6)=C(10,6)\cdot 0.3125^6\cdot (1-0.3125)^{10-6}=\frac{10!}{6!4!}\cdot 0.3125^6\cdot 0.6875^4=$

$=0.0437$

b) $P(X\ge5)=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)$

$P(X=5)=C(10,5)\cdot 0.3125^5\cdot 0.6875^5=0.1153$

$P(X=6)=0.0437$

$P(X=7)=C(10,7)\cdot 0.3125^7\cdot 0.6875^3=0.0113$

$P(X=8)=C(10,8)\cdot 0.3125^8\cdot 0.6875^2=0.0019$

$P(X=9)=C(10,9)\cdot 0.3125^9\cdot 0.6875^1=0.0002$

$P(X=10)=C(10,10)\cdot 0.3125^10\cdot 0.6875^0=0.000009$

$P(X\ge5)=0.1153+0.0437+0.0113+0.0019+0.0002+0.000009=0.1724$

c)

$P(X\le2)=P(X=0)+P(X=1)+P(X=2)$

$P(X=0)=C(10,0)\cdot 0.3125^0\cdot 0.6875^10=0.0236$

$P(X=1)=C(10,1)\cdot 0.3125^1\cdot 0.6875^9=0.1072$

$P(X=2)=C(10,2)\cdot 0.3125^2\cdot 0.6875^8=0.2193$

$P(X\le2)=0.0236+0.1072+0.2193=0.3501$

d)

$E(X)=np=10\cdot0.3125=3.125\simeq4$

The expected number of laptop recipients who do not have Internet access is 4.

e)

$E(X)=3.125$

$var(X)=np(1-p)=10\cdot0.3125\cdot0.6875=2.1484$

$sd(X)=\sqrt{var(X)}=\sqrt{2.1484}=1.4657$

Answer:

a) 0.0437

b) 0.1724

c) 0.3501

d) The expected number of laptop recipients who do not have Internet access is 4.

e) 3.125, 2.1484, 1.4657.