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View Full DocumentExam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Determine whether the given procedure results in a binomial distribution. If not, state the reason why. 1) Spinning a roulette wheel 6 times, keeping track of the occurrences of a winning number of "16". 1) A) Procedure results in a binomial distribution.. B) Not binomial: there are more than two outcomes for each trial. C) Not binomial: the trials are not independent. D) Not binomial: there are too many trials. Solve the problem. 2) On a multiple choice test with 18 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the variance for the number of correct answers. 2) A) 33.8 B) 11.4 C) 1.8 D) 3.4 Find the indicated mean. 3) A certain rare form of cancer occurs in 32 children in a million, so its probability is 0.000032. In the city of Normalville there are 76,430,000 children. A Poisson distribution will be used to approximate the probability that the number of cases of the disease in Normalville children is more than 2. Find the mean of the appropriate Poisson distribution (the mean number of cases in groups of 76,430,000 children). 3) A) 245 B) 2450 C) 24,500 D) 0.000032 Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. 4) Shaded area is 0.0694. 4) A) 1.45 B) 1.39 C) 1.26 D) 1.48 Solve the problem. Round to the nearest tenth unless indicated otherwise. 5) The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.3 and a standard deviation of 40.4. All units are in mg/100 mL. Find the two levels that separate the top 9% and the bottom 9%. 5) A) 124.2 mg/100mL and 232.4 mg/100mL B) 108.0 mg/100mL and 248.6 mg/100mL C) 165.4 mg/100mL and 191.23 mg/100mL D) 161.7 mg/100mL and 194.9 mg/100mL Solve the problem. 6) Human body temperatures are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. If 19 people are randomly selected, find the probability that their mean body temperature will be less than 98.50F. 6) A) 0.4826 B) 0.3343 C) 0.0833 D) 0.9826 1 Find the indicated critical z value. 7) Find the critical value z /2 that corresponds to a 98% confidence level. 7) A) 1.75 B) 2.575 C) 2.33 D) 2.05 Do one of the following, as appropriate: (a) Find the critical value z /2 , (b) find the critical value t /2 , (c) state that neither the normal nor the t distribution applies. 8) 95%; n = 11; is known; population appears to be very skewed. 8) A) z /2 = 1.96 B) z /2 = 1.812 C) t /2 = 2.228 D) Neither the normal nor the t distribution applies.... View Full Document
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