7.3 problems

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Sample Means 60. A. assuming the machine is working properly and “normal” the population distribution is N(40.125, 0.002). The mean of the sampling distribution is the same as the original population distribution so 40.125. B. the standard deviation is 0.002/ sqrt(4) = 0.001mm 4(10)= 40, since we are not directly told what the population is we can safely assume that it is above 40 axles in an hour if the machine is working properly. 62. We would need a sample size of 16 to get a standard deviation of 0.0005mm. This is because the square root of 16 is 4 and if we divide the original population distribution deviation by 4 we would get 0.0005mm. 63. A. 50(10) = 500. If we were trying to find a random sample of students from a given population and chose 50 students, the population would have to have been 500. The maximum number of students we would be able to take from a population of 200 would’ve been 20. B. Violating the 10% condition can cause an unrepresentative standard deviation of the population, decreasing the variability so much so that it is no longer an accurate representation of the population. 65. State: probability that a randomly selected bottle contains less than 295ml Plan: already states the population is randomly selected and a normal distribution Do: N(298, 3) Z295 = 295-298/3= -1 -1 on table A is = 0.1587 Conclude: The probability that a randomly selected bottle contains less than 295ml of cola is 0.1587. 67. A. we are told it is a normal distribution = N(188, 41). If we took an SRS of 100 men from the population the sampling distribution would also be approximately normally distributed as well. B. Finding the new standard deviation of the sampling distribution is as follows: 41/ sqrt(100)= 4.1 P (185≤x≤191) = Z185 = 185-188/4.1 = -0.731 =0.2327 Z191 = 191-188/4.1 = 0.731 = 0.7673 0.7673-0.2327= 0.5346 The probability that x(bar) estimates within plus or minus 3 mg/dl is 0.5346. C. 41/sqrt (1000) = 1.2965

Z185 = 185-188/1.2965 = -0.731 =-2.313 = 0.0104 Z191 = 191-188/1.2965 = 0.731 = 2.313 = 0.9896 0.9896-0.0104 = 0.9792 The probability that x(bar) estimates within plus or minus 3 mg/dl is 0.9792 which is a lot higher of a probability than with a sample size of 100. The larger sample is “better” in giving estimates of percentages of populations that lie in between two points. 69. A. N (48,8.2) 8.2/sqrt(8)= 2.899 z42.2= 42.2-48/2.899 = -2.00 = 0.0228 The probability that the sample mean lifetime is 42.2 months or less is approximately 0.0228. B. There is not convincing evidence that the company is overstating the average lifetime of its batteries, seeing as not a large proportion of lifetimes lie at 42.2 months or less at all. 71. A. The shape of the sampling distribution for an srs of size n = 5 will also likely be skewed to the right with a very similar distribution as the population distribution we were originally given because such a small sample size does not contribute the the central limit theorem. B. According to the central limit theorem a population size significant enough (in this case 100 is n≥30) can make a mean sampling distribution reﬂect a normal distribution, in this case it would be unimodal, roughly symmetric, with no significant outliers. 74. A. State: what is the probability that customers grab an average of $15 or more. Plan: randomized- a simple random sample is taken 10% - we can safely assume that 10% of the players over time exceeds 400 Do: N (13,9) for the population distribution Std. dev.= 9/ sqrt(40) = 1.423 Z15 = 15-13/1.423 = 1.405 =1- 0.919 = 0.081 The probability of customers grabbing an average of 15 dollars or more is approximately .081 which is extremely unlikely. B. z? = ?-13/1.423 = 1.64 $15.33 could allow them to be 95% certain the mean amount grabbed was less than what they charged. 76. A. randomly selecting one car entering this interchange during rush hour and finding two or more people in it is more likely because if you took 35 cars, the average would end

up being closer to 1.6 people per car since you would have to find more cars that are considered “outliers” or unusual to the population because they have more people than what the distribution is stating. B. We can’t use a normal distribution because it would not accurately represent just one car, it would have to be a sample population of 30 or more according ot the central limit theorem. C. N(1.6,0.75) 0.75/ sqrt(35) = 0.1267 Z2 = 2-1.6/0.1267 = 3.157 = 0.0008 The probability of getting 35 cars entering during rush hour with two or more people is 0.0008, or very unlikely. 77 . No, because the central limit theorem deals only with the distribution of sample means and not with the distribution of histograms. 78. Yes, because the central limit theorem describes that with an increase of sample size is a decrease in variability since the normal distribution becomes more and more concise. 80. State: what is the probability that the total number of lightning strikes is less than 250? Plan: randomized- states random sample 10% - 250(10) = 2500 lightning strikes is reasonable to conclude occur in a given year Large numbers - 250 is greater than or equal to 30 Do: N(6, 2.4) 50 x 6 = 300 50 x 2.4 = 120 N (300,120) Z250 = 250-300/120 = -0.416 = 0.3372 The probability that the total number of lightning strikes in a random sample of 50 square-kilometer plots of land is less than 250 is 0.3372. 82. A. NFL(26.2, 3.24) NBA(25.8,4.24) 26.2 - 25.8 = -0.6 B. 3.24/sqrt(20)= 0.7244 4.24/ sqrt(20)=0.948 20(10)=200 we can reasonably conclude that there are more than 200 players in either then NFL and the NBA.

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C. No, the shape of the distribution is likely still skewed right because the sample size was not large enough to qualify for the central limit theorem. 87. B 88. A 89. E z?= ? - 16.05/0.1 = 1.64 = 16.21 90. D

7.3 problems

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