HW 310

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Apr 3, 2024

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Question 1: a) To find how many standard deviations away 5'10" is from the mean, you can use the formula: (number - mean) / standard deviation (5'10" - 5'8") / 2" (70" - 68") / 2" 2" / 2" = 1 standard deviation b) For 5'11": (5'11" - 5'8") / 2" (71" - 68") / 2" 3" / 2" = 1.5 standard deviations c) The z-score for 5'10" is the same as the number of standard deviations, so it is 1. d) The z-score for 5'11" is the same as the number of standard deviations, so it is 1.5. e) The previous four answers show that z-scores and standard deviations are equivalent in this context. The z-score is the number of standard deviations away from the mean. Question 2: a) To find the probability that there will be less than 27,500 people at a game using Norm.Dist(): =NORM.DIST(27500, 30000, 5000, TRUE) The answer is approximately 0.3085. b) To find the probability for a realization less than z = -0.5 using Norm.S.Dist(): =NORM.S.DIST(-0.5, TRUE) The answer is approximately 0.3085. c) To find the probability there will be more than the mean plus two standard deviations of guests using Norm.Dist(): =NORM.DIST(30000 + 2 * 5000, 30000, 5000, TRUE) =1 - NORM.DIST(40000, 30000, 5000, TRUE) The answer is approximately 0.0228. d) To find the probability using Norm.S.Dist(): =1 - NORM.S.DIST(2, TRUE) The answer is approximately 0.0228. Question 3: a) Using the empirical rule, the probability of being within two standard deviations of the mean in a normal distribution is approximately 95%. b) The probability of being more than two standard deviations larger than the mean in a normal distribution is approximately 2.5%. c) To find the probability of being within 1.7 standard deviations of the mean using Norm.S.Dist(): =NORM.S.DIST(1.7, TRUE) - NORM.S.DIST(-1.7, TRUE) The answer is approximately 0.9115.

d) To find the probability of being within 0.9 standard deviations of the mean using Norm.S.Dist(): =NORM.S.DIST(0.9, TRUE) - NORM.S.DIST(-0.9, TRUE) The answer is approximately 0.6283. Question 1: a) As stated by the central limit theorem, when you take a sample mean from a population, it is pulled from a normal distribution. b) The sampling distribution has the same mean as the population mean. c) The sampling distribution has a standard deviation equal to the population standard deviation divided by the square root of the sample size (known as the standard error). d) In this case, the mean weight of cockatoos is 15 lbs, so the mean of the sampling distribution is also 15 lbs. The standard deviation of the sampling distribution is the population standard deviation (3 lbs) divided by the square root of the sample size (100): 3 / √100 = 3 / 10 = 0.3 lbs. e) To find the probability that your sample would have an average greater than 15.05 lbs, you can calculate the z-score for the sample mean and use the normal distribution. Z = (sample mean - population mean) / standard error Z = (15.05 - 15) / 0.3 = 0.05 / 0.3 ≈ 0.167 Using a z-table or a calculator, the probability of a z-score greater than 0.167 is approximately 0.4335. f) To find the probability that a single cockatoo would have a weight greater than 15.05 lbs, use the z-score for the population distribution: Z = (15.05 - 15) / 3 = 0.05 / 3 ≈ 0.0167 Using a z-table or a calculator, the probability of a z-score greater than 0.0167 is approximately 0.4934. Question 2: a) For this situation, you would use the sampling distribution of the sample mean, which is a normal distribution. This is because of the central limit theorem, which states that the distribution of sample means will be approximately normally distributed when the sample size is large. b) To create a 95% confidence interval for the sample average of cockatoo weights, use the following formula: (sample mean) ± (z-score for 95% confidence) * (standard error) In this case: 15 ± (1.96) * (0.3) 15 ± 0.588 The 95% confidence interval is (14.412, 15.588) lbs. Question 3: a) The null hypothesis (H0) is that the mean weight of a Galapagos Island gecko is no more than 15.8 grams (µ ≤ 15.8). The alternative hypothesis (H1) is that the mean weight is greater than 15.8 grams (µ > 15.8).

b) To calculate the z test statistic, use the formula: Z = (sample mean - null hypothesis mean) / (standard deviation / √sample size) Z = (16.1 - 15.8) / (1 / √40) = 0.3 / (1 / √40) ≈ 1.897 c) For a 95% confidence interval on a right-tailed test, the z-score is approximately 1.645. Since the calculated z-score (1.897) is greater than the critical z-score (1.645), we would reject Darwin's claim. d) To find the p-value associated with the sample, look up the probability of a z-score greater than 1.897 in a z-table or use a calculator. The p-value is approximately 0.0287. This is less than the alpha level of 0.05, so we would reject Darwin's claim based on this

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