EXAM #2 Problem -1 a. 0.0401 b. .1587 c. .1359 d. 1.04 e. -1.64 f. 0.0455 g. .9599 h. .9332 i. 1.00 and -1.00Problem -2 a. .1587 b. .0228 c. .3812 d. .1818 e. 51.2667 f. 43.5922 g. -.0600 and .0600 h. .4756 i. 1 Problem -3. a. .2266 b. .3085 c. .6827 d. 0455 e. 102.6980 f. 98.9866 g. 97.90 and 102.10 h. .6827 i. 30Problem -4. a. 675.00 b. The sampling distribution is normally distributed because there is a large sample size (>30) so it it follows the Central Limit Theorem . c. 40429.32 d. 43906.68 e. yes f. no Problem -5. a. 43.04 b. 41.9260 c. 50 d. 5.9292 e. The sampling distribution is normally distributed because there is a large sample size (>30) so it it follows the Central Limit Theorem . f. 31.12 g. 54.96 h. yes i. no Problem -6.Ho – u=1; (The mean amount of paint is 1 gallon); H1 – u=/=1 (The mean amount of paint is not 1 gallon) a. Decision Rule states to reject Ho if lZstatl > 2.5758 b. test statistic: Zstat = (xBar – u)/(sd/n) = (.995 – 1) / (.02/sqrt 50) = -1.7678 c. 2.5758 d. .0771 e. lZstatl282; (The mean driving distance is greater than 282 yards); H1 – u1.6766. There is sufficient evidence to suggest that the mean driving distance is not greater than 282 h. p-value = .0001 2.0017 or p-value 2.0017.There is sufficient to conclude that there is a difference in the means of the average spent on couples vacationing in Jamaica versus Bermudah. p-value = .0001