Answer to Question #262330 in Statistics and Probability for hammad

Question

. Fifty percent of Americans believed the country was in a recession, even though technically the economy had not shown two straight quarters of negative growth (BusinessWeek, July 30, 2001). For a sample of 20 Americans, make the following calculations. a. Compute the probability that exactly 12 people believed the country was in a recession. b. Compute the probability that no more than five people believed the country was in a recession. c. How many people would you expect to say the country was in a recession? d. Compute the variance and standard deviation of the number of people who believed the country was in a recession.

Answer 1

Let X be the amount of american from sample of 20 people who believe country was in a recession.

Then X ~ Bin(20, 0.5)

(a) $P(X=12)={20 \choose 12}*(0.5)^{12}*0.5^8=0.1201$

(b) $P(X≤5)=P(X=0)+...+P(X=5)={20 \choose 0}*0.5^{20}+...+{20 \choose 5}*0.5^{20}=0.0000+0.0000+0.0002+0.0011+0.0046+0.0148=0.0207$

(c) We have to find the expected value of X

$E(X)=np=20*0.5=10$ people we should expect

(d) $D(X)=np(1-p)=20*0.5*0.5=5$ is the variance;

$\sigma(X)=\sqrt{D}=\sqrt{5}$ is the standard deviation.