Answer to Question #169739 in Statistics and Probability for Miu

Question

A survey by Frank N.Magid Associates revealed that 3% of Americans are not connected to the Internet at home. Another researcher randomly selects 70 Americans.

a. What is the expected number of these who would not be connected to the Internet at home?

b. What is the probability that 8 or more are not connected to the Internet at home?

c. What is the probability that between 3 and 6 (inclusive) are not connected to the Internet at home?

Appendix A Statistical Tables

*(Round your answer to 1 decimal place.)

**(Round your answers to 4 decimal places when calculating using Table A.3.)

a. Expected number = 

enter the expected number rounded to 1 decimal place

 ^*

b. P(x ≥ 8) = 

enter probability rounded to 4 decimal places

 ^**

c. P(3 ≤ x ≤ 6) = 

enter probability rounded to 4 decimal places

 ^**

The Table A.3 is "Poisson Probability"

Answer 1

Let $X=$ the number of people  not connected to the Internet at home

$n=70, p=0.03$

Binomial approximation to Poisson: $\lambda=np=70(0.03)=2.1$

a) $E(X)=\lambda=2.1$

b)

$$P(X\geq8)=1-P(X<8)=0.0015$$

c)

$$P(3\leq X\leq6)=P(X\leq6)-P(X<3)$$

$$=0.99414-0.64963=0.3445$$