Question

1)

The mean per capita income is 21,053 dollars per annum with a standard deviation of 805 dollars per annum.

What is the probability that the sample mean would be less than  20876 dollars if a sample of 71 persons is randomly selected? Round your answer to four decimal places.

2) Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.

P(X<5), n=8, p=0.4

3)

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 28,631miles, with a variance of 16,834,610

What is the probability that the sample mean would be less than 28,455 miles in a sample of 178 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

1)

P(x< 20876)
= P(z< ( x -mean)/(sigma/sqrt(n))
= P(z < ( 20876 - 21053)/(805/sqrt(71))
= P(z < -1.85)
= 0.0320

2)

As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

P(x< 5) = 8C0 * 0.4^0 * (1-0.4)^8 + 8C1 * 0.4^1* (1-0.4)^7 + 8C2 * 0.4^2 * (1-0.4)^6 + 8C3 * 0.4^3 * (1-0.4)^5 + 8C4 * 0.4^4 * (1-0.4)^4

P(x< 5) = 0.8263

3)

P(x<28455)
= P(z< ( x -mean)/(sigma/sqrt(n))
= P(z< ( 28455 - 28631)/(4103.0001/sqrt(178))
= P(z< -0.57
= 0.2836