question 1. In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation.

A. If the table is a probability distribution, what is its mean? Select the correct choice below and fill in any answer boxes within your choice

b. If the table is a probability distribution, what is its standard deviation? Select the correct choice below and fill in any answer boxes within your choice.

Question 2

Determine whether or not the procedure described below results in a binomial distribution. If it is not binomial, identify at least one requirement that is not satisfied.Five hundred different voters in a region with two major political parties, A and B, are randomly selected from the population of 5.2 million registered voters. Each is asked if he or she is a member of political party A, recording Yes or No. Choose the correct answer below.

A. No, the probability of success is not the same in all trials.

B.No,thetrialsarenotindependentandthesampleismorethan5%ofthepopulation.

C.Yes,theresultisabinomialprobabilitydistribution.

D. No, the number of trials is not fixed.

E. No, there are more than two possible outcomes.

Question 3.

Assume the readings on thermometers are normally distributed with a mean of 0C and a standard deviation of 1.00C. Find the probability that a randomly selected thermometer reads between 2.15 and 0.35 and draw a sketch of the region.

Draw a sketch and probability is ? (Round to four decimal places as needed.)

question 4

If np 5 and nq 5, estimate P(morethan8) with n=11 and p=0.7 by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq <5, then state that the normal approximation is not suitable

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

A. P(more than 8)= (Round to four decimal places as needed.)

B. The normal distribution cannot be used.