Math 251, Test 2 Wednesday, May 19, 2004 Name: Hints and Answers Instructions. Complete each of the following 9 problems. Please show all appropriate details in your solutions. Good Luck. 1. (15 pts) (a) Find the mean and standard deviation for the sampling distribution for x based on a sample size of 81 from a population with mean 37 and standard deviation 6. Answer. µ x = 37 and σ x = 6 81 = 2 3. (b) Find z c when c =.96. Answer. Find z-value corresponding to an area of.5 +.48/2 =.98. Thus z.96 2.05. (c) What z-value would you obtain for a data value that is 3 standard deviations below the mean? Answer. z = 3. (d) If a population has a standard deviation of 10, what sample size would be necessary in order for a 99% confidence interval to estimate the population mean within ±2? Answer. n = ( zc σ E ) ( ) 2 2 2.58 10 = 166.41. Thus we would use a sample size of n = 167. 2 (e) What size of random sample is needed by a polling organization to estimate a population percentage within plus or minus 1% with 95% confidence. In your calculation assume that there is no preliminary estimate for p. Answer. n = 1 4 ( ) 2 1.96 = 9604..01 (f) What proportion of data in a normal population lies within 2.5 standard deviations of the mean? Answer. From the table, we find P ( 2.5 < z < 2.5) =.9938.0062 =.9876. (g) Let x be the random variable that represents the weight of a randomly selected Gabon Viper. Is x a continuous or discrete random variable? Answer. x is a continuous random variable, because it can theoretically take any value larger than 0, i.e. it can be any value in an interval.
2. (a) (1 pt) Fill in the missing probability for the following discrete random variable. x 3 4 5 7 8 10 P (x).2.3.1.2.15.05 (b) Find the mean and standard deviation of the random variable x with distribution as follows. x 0 2 3 5 P (x).20.35.30.15 Answer. The mean is µ = xp (x) = 2(.35) + 3(.30) + 5(.15) = 2.35. While σ 2 = 2 2 (.35) + 3 2 (.30) + 5 2 (.15) 2.34 2 = 2.3275. Therefore, the standard deviation is σ = 2.3275 1.5256. 3. Suppose the probability Kobe Bryant will make a free throw is 0.84. Suppose Mr. Bryant will have 14 free throw attempts in his next game. Find: (a) (2 pts) The probability that he will make all 14 free throws. Answer. P (r = 14) =.84 14 =.087078. (b) (2 pts) The probability that he will make exactly 13 free throws. Answer. P (r = 13) = C 14,13 (.84) 13 (.16) 1 = 14(.84) 13 (.16) =.232208822685. (c) (1 pts) The probability that he will make 12 or fewer free throws. Answer. The answer is approximately 1 (.0871 +.2322) = 1.3193 =.6807.
4. Northwest Airlines has found that 96% of people with tickets will show up for their Friday afternoon flight from Detroit to Amsterdam. Suppose that there are 300 passengers holding tickets for this flight, and the jet can carry 294 passengers, and that the decisions of passengers to show up are independent of one another. (a) (1 pts) Verify that the normal approximation of the binomial distribution can be used for this problem. Answer. We check that np = (300)(.96) = 288 > 5 and nq = (300)(.04) = 12 > 5. (b) (4 pts) Use the normal approximation to the binomial distribution to find the probability that from 285 to 294 passengers (including 285 and 294) will show up for the flight. Answer. By (a), this binomial distribution is approximately normal, and it has mean µ = 288 and standard deviation σ = npq = (300)(.96)(.04) 3.394. Using the continuity correction, we compute P (285 r 294) P (284.5 x 294.5) = P ( 1.03 z 1.915) =.9726.1515 =.8211. 5. Suppose the distribution of weights of adult female American Landrace pigs is normally distributed with a mean of 300 lbs and standard deviation of 40 lbs. (a) (2 pts) What weight is at the 20th percentile? Answer. The z-value corresponding to the 20th percentile is z =.84, thus the corresponding weight is x = 300 + (.84)(40) = 266.4 lbs. (b) (2 pts) What proportion of adult female American Landrace pigs weigh between 230 lbs and 290 lbs? Answer. P (230 < x < 290) = P ( 1.75 < z <.25) =.4013.0401 =.3612. (c) (1 pt) What proportion of adult female American Landrace pigs weigh more than 290 lbs? Answer. P (x > 290) = P (z >.25) = 1.4013 =.5987.
6. (5 pts) The encyclopedia of erroneous data reports that the mean weight of adult male polar bears is 1100 lbs. (a) Suppose a wildlife biologist collects a random sample of 64 adult male polar bears and finds that x = 1150 lbs and s = 130 lbs. If the true population mean is 1100 lbs, what is the probability of finding a sample mean of x = 1150 lbs or greater? Assume that σ = 130 lbs. Answer. 1.9990 =.001. P ( x 1150) = P ( z ) 1150 1100 130 8 = P (z 3.077) 1 P (z < 3.08) = (b) Based on the sample in (a), would you be skeptical of that the mean weight of adult male polar bears is 1100 lbs? Explain. Answer. Very skeptical. If the mean weight were 1100 lbs, there would only be.001 probability of obtaining the sample in (a), hence, assuming the biologists sample was random, and the statistics are correct, we are quite certain that the true mean weight is more than 1100 lbs. 7. (5 pts) In a recent poll of 1009 randomly selected adult Americans it was reported that 44% of adult Americans surveyed approve of the way the United States has handled the war in Iraq. (a) Find a 95% confidence interval for the proportion of adult Americans that approve of the way the United States has handled the war in Iraq. (.44)(.56) Answer. The confidence interval has endpoints.44 ± 1.96, and so the interval is 1009 (.4094,.4706). (b) Based on (a) would you be comfortable saying that less than 45% of all adult Americans approve of the way the United States has handled the war in Iraq? Explain. Answer. No, you should not be comfortable saying that because you are 95% sure that the true percentage of support is between 40.94% and 47.06%; which leaves a good portion of interval above 45%, indicating there is a reasonable chance that the true level of support is more than 45%.
8. (5 pts) A recent survey of 400 randomly selected gas stations in America found that the average price for unleaded gasoline is $2.02. That is, in their sample, they found x = $2.02 and s = $0.15. (a) Find a 99% confidence interval for the mean price of unleaded gasoline in America. Answer. The confidence interval has endpoints 2.02 ± 2.58.15 400. Thus the 99% confidence interval is (2.00065, 2.03935). (b) Based on your interval in (a), would you be comfortable saying that the mean price of unleaded gas in America is at least $2.00? Explain. Answer. Yes, I am 99% sure that the true mean price is between $2.00 and $2.04 which is at least $2.00. 9. (5 pts) In a random sample of 50 gas stations in California it was found that the sample mean price of unleaded gasoline was $2.21 with a standard deviation of $0.17. At the same time, a random sample of 60 gas stations in New York had a sample mean price of $2.17 for unleaded gasoline with a standard deviation of $0.16. (a) Find a 95% confidence interval for the difference in unleaded gasoline prices between California and New York. (.17) 2 Answer. The confidence interval has endpoints (2.21 2.17) ± 1.96 + (.16)2 50 60, so the endpoints are approximately.04 ± (1.96)(.031696). Hence the confidence interval is (.0221,.1021). (b) Interpret the interval in (a). Does it strongly suggest that gasoline prices in California are higher than in New York? Answer. We are 95% sure that the mean price of unleaded gasoline in California is from 2 cents lower to 10 cents higher than the mean price in New York. Hence there is still a reasonable possibility that the mean price in California is lower, therefore the interval in (a) does not strongly suggest that the mean price in California is higher.