Question 1 (3 points) Assume that a random variable X comes from a Poisson distribution with parameter lambda. If we wanted to use the normal approximation to the Poisson to find probabilities, the mean used in the normal distribution would be O A O O O VA Question 2 (3 points) 1. A cost accountant wished to establish the average amount spent by executives per day on travel and lodging. A random sample of 50 executive expense receipts is taken. The average in the sample was $208, and the standard deviation was $29. The accountant prepares a 90-percent confidence interval, which is given below: 208 Mean StDev SE Mean 904 CI 50 208 .00 29 .00 4.10 (201 .12, 214.88] The interpretation of this interval is the probability that the true mean is in the given interval is 0 since the sample mean is in the interval, the probability that the true mean is in there is 1 the probability that the true mean is in the given interval is 90% Owe are 90% confident that the true mean is between 201.12 and 214.88Question 3 (3 points) 1. A cost accountant wished to establish the average amount spent by executives per day on travel and lodging. A random sample of 50 executive expense receipts is taken. The average in the sample was $208, and the standard deviation was $29. The accountant prepares a 90-percent confidence interval, which is given below: 208 NJ Mean StDev SE Mean 90% CI 50 208.00 29.00 4.10 (201 .12, 214. 88] This distribution was created using a t-critical value. What additional information do we need to show that the t-distribution was the correct choice? O the degrees of freedom the sample was taken from a normal distribution the value for the population standard deviation O we have all of the information that we need since the standard deviation is given Question 4 (3 points) 1. A cost accountant wished to establish the average amount spent by executives per day on travel and lodging. A random sample of 30 executive expense receipts is taken. The average in the sample was $208, and the standard deviation was $29. The accountant prepares a 90-percent confidence interval, which is given below: 208 NJ Mean StDev SE Mean 904 CI 50 208 .00 29.00 4.10 (201 .12, 214.88) What would happen to the interval if we increased the confidence level to 95%? the interval would become more narrow since we need more values to fit inside O the interval would widen since we need more values to fit inside the interval would become more narrow since we need less values to fit inside the interval would widen since we need less values to fit inside