Recall that a customer is considered to be very satisfied with his or her XYZ Box video game system if the customer's composite score on the survey instrument is at least 42. One way to show that customers are typically very satisfied is to show that the mean of the population of all satisfaction ratings is at least 42. Letting this mean be /, in this exercise we wish to investigate whether the sample of 65 satisfaction ratings provides evidence to support the claim that / exceeds 42 (and, therefore, is at least 42). For the sake of argument, we begin by assuming that / equals 42, and we then attempt to use the sample to contradict this assumption in favor of the conclusion that / exceeds 42. Recall that the mean of the sample of 65 satisfaction ratings is x = 42.95, and assume that O, the standard deviation of the population of all satisfaction ratings, is known to be 2.64. Click here for the Excel Data File (a) Consider the sampling distribution of x for random samples of 65 customer satisfaction ratings. Use the properties of this sampling distribution to find the probability of observing a sample mean greater than or equal to 42.95 when we assume that / equals 42. (Round O ; and p to 4 decimal places. Round z-scores to 2 decimal places. Do not round your intermediate calculations.) O = UX = OX = P(x > 42.95) = (b) If u equal to 42, what percentage of all possible sample means are greater than or equal to 42.95? Since we have actually observed a sample mean of x = 42.95, is it more reasonable to believe that (1) u equal to 42 and we have observed a sample mean that is greater than or equal to 42.95 when equal to 42, or (2) that we have observed a sample mean that is greater than or equal to 42.95 because is greater than 42? Explain. What do you conclude about whether customers are typically very satisfied with the XYZ Box video game system? (Round your answer to 2 decimal places. Do not round your intermediate calculations. Input your answer to percent without percent sign.) %; conclude 42.