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STAT_12A_Practice Score: 3.6/12 3/6 answered Question 2 Score on last try: 0.6 of 1 pts. See Details for more. > Next question You can retry this question below Describe the sampling distribution (mean, standard deviation, and shape) of the mean weight of the passengers plus the luggage for a random sample of 28 customers. Fill in the blanks. Since the population distribution of the weight of the passengers plus the luggage is approximately normal | of , the sampling distribution of the mean weight of the passengers plus the luggage for a random sample of 25 x customers will also be approximately normal V o , with a mean of 210 v o pounds and a standard deviation of 25 x pounds (rounded to the hundredths place). The following is a sketch of the sampling distribution described in Question 2 superimposed onto the distribution of weight of passengers plus luggage described in Question 1. The total area under the curve for each distribution equals 1; the height of the distribution of weight of passengers plus luggage is lower, which means it has greater variability. Individual weight Mean weight 135 160 185 210 235 260 285 Passenger plus luggage weight (pounds)STAT_12A_Practice Score: 3.6/12 3/6 answered X Question 3 Score on last try: 0 of 1 pts. See Details for more. > Next question You can retry this question below Assume that passengers on any particular flight are similar to a random sample of all passengers. If the total weight of the passengers and the luggage should not exceed 6,160 pounds, what is the probability that a sold-out flight (28 passengers and their luggage) will exceed the weight limit? Round to four decimal places. Hint Rewrite the desired limit as an average per passenger. Using the DCMP Normal Distribution tool, calculate the probability that the mean weight is greater than 6160/28 = 220 pounds using the mean and standard deviation from Question 2: The Normal Distribution Explore F Find Probability Find Percentile/Quantile Normal Distribution with u = 210 and o = 4.725 Mean H P(X > 220) = 1.72% 210 98.28%, 1.72% Standard Deviation o: 4.725 Type of Probability: Upper tail: P(X > x) Value of x: 220 195.825 200.550 205.275 210.000 214.725 219.450 224.175 220 Download Graph Normal Probability (Upper Tail): X P(X x) 210 4.7250 220 1.716% 1.72 X Calculator Submit QuestionSTAT_1 2A_Practice Score: 3.6/12 3/6 answered 0 Question 4 v Questions 46: In In-Class Activity 2A, you practiced ta king random samples from a population of words from an excerpt of a 2001 speech by Supreme CourtJustice Sonia Sotomayorr which was given at the University of California at Berkeley on October 26, 2001.1 The word length (number of letters) of each word was measured. The mean word length in the entire speech was 4.68 letters and the standard deviation was 2.61 letters. Go to the DCMP Sampling Distribution of the Sample Mean {Discrete Population) tool at Under "Select Population Distribution,\"select "Word Length Sotomayor." A plot of this population distribution of the word lengths will be shown at the top of the page and is displayed below. Note that the population distribution of the word lengths is skewed right. Population Distribution p = 4.68. o = 2.61 0.2' P(X=x) 0,1- o_0..A...-q-.q l 2 3 4 5 6 7 8 9 10 'll 12 13 14 15 'IB 1Berkley Law. (2009, May 26). Supreme Court Nominee Sonia Sotomayor's speech at Berkley Low in 2001. mpszllwww.|aw.berkeley.edularticlelsupreme-court-nominee-sonia-sotomayors-speech-at-berkeley-law in-2001l Score on last try: 1 of 3 pts. See Details for more. > Next question You can retry this question below Set the sample size to 5 and generate 1,000 random samples of this size. Select the boxes for "Show Normal Approximation" and "Find Probability." Berkley Law. (2009, May 26). Supreme Court Nominee Sonia Sotomayor's speech at Berkley Law in 2001. https://www.law.berkeley.edu/article/supreme-court-nominee-sonia-sotomayors-speech-at-berkeley-law- in-2001/ Score on last try: 1 of 3 pts. See Details for more. > Next question You can retry this question below Set the sample size to 5 and generate 1,000 random samples of this size. Select the boxes for "Show Normal Approximation" and "Find Probability." Part A: What proportion of simulated mean word lengths is less than 2.5 characters? Part B: Calculate the theoretical mean and standard deviation of the sampling distribution of the sample means for samples of size 5. Using the DCMP Normal Distribution tool https://dcmathpathways.shinyapps.io/NormalDist/, find the probability of a 2.5 or less average word length on a normal distribution with this mean and standard deviation. Round to three decimal places. = Standard Deviation = Probability (2.5 or less) =STAT_12A_Practice Score: 3.6/12 3/6 answered . Question 5 Questions 4-6: In In-Class Activity 2.A, you practiced taking random samples from a population of words from an excerpt of a 2001 speech by Supreme Court Justice Sonia Sotomayor, which was given at the University of California at Berkeley on October 26, 2001. The word length (number of letters) of each word was measured. The mean word length in the entire speech was 4.68 letters and the standard deviation was 2.61 letters Go to the DCMP Sampling Distribution of the Sample Mean (Discrete Population) tool at Under "Select Population Distribution," select "Word Length Sotomayor." A plot of this population distribution of the word lengths will be shown at the top of the page and is displayed below. Note that the population distribution of the word lengths is skewed right. Population Distribution H = 4.68, 0 = 2.61 0.2 P(X=X) 0.1 - 0.0 5 8 9 10 11 12 13 14 16 Berkley Law. (2009, May 26). Supreme Court Nominee Sonia Sotomayor's speech at Berkley Law in 2001. https://www.law.berkeley.edu/article/supreme-court-nominee-sonia-sotomayors-speech-at-berkeley-law- in-2001/ Score on last try: 1 of 3 pts. See Details for more. > Next question You can retry this question below Go back to the DCMP Sampling Distribution of the Sample Mean (Discrete Population) tool at https://dcmathpathways.shinyapps.io/SampDist discrete/ . Under "Select Population Distribution," select "Word Length Sotomayor."Score on last try: 1 of3 pts. See Details for more. > Next question You can retry this question below Go back to the DCMP Sampling Distribution of the Sample Mean (Discrete Popuiation) tool at msz/dcmathpathwaysshinygpszo/SampDist discrete] . Under\"Select Population Distribution," select "Word Length Sotomayor.\" Set the sample size to 50 and generate 1,000 random samples of this size. Select the boxes for \"Show Normal Approximation" and "Find Probability." Part A: What proportion of simulated mean word lengths is larger than 5.3 letters? Part B: Calculate the theoretical mean and standard deviation of the sampling distribution of the sample means for samples of size n : 50. Using the DCMP Normall Distribution tool Epszlldcmathpathwaysshinygpszo/NormalDist/ , find the probability of a 5.3 or larger on a normal distribution with this mean and standard deviation. Round to three decimal places. Mean : Standard Deviation : Probability (5.3 or larger) 2 STAT_12A_Practice Score: 3.6/12 3/6 answered Question 6 Questions 4-6: In In-Class Activity 2.A, you practiced taking random samples from a population of words from an excerpt of a 2001 speech by Supreme Court Justice Sonia Sotomayor, which was given at the University of California at Berkeley on October 26, 2001. The word length (number of letters) of each word was measured. The mean word length in the entire speech was 4.68 letters and the standard deviation was 2.61 letters. Go to the DCMP Sampling Distribution of the Sample Mean (Discrete Population) tool at Under "Select Population Distribution," select "Word Length Sotomayor." A plot of this population distribution of the word lengths will be shown at the top of the page and is displayed below. Note that the population distribution of the word lengths is skewed right. Population Distribution H = 4.68, 0 = 2.61 0.2 P(X=X) 0.1 0.0 2 5 6 7 8 9 10 11 12 13 14 15 16 Berkley Law. (2009, May 26). Supreme Court Nominee Sonia Sotomayor's speech at Berkley Law in 2001. https://www.law.berkeley.edu/article/supreme-court-nominee-sonia-sotomayors-speech-at-berkeley-law- in-2001/ Historians are interested in comparing the word lengths for speeches by Justice Sonia Sotomayor to speeches by another Supreme Court Justice. They take a random sample of 50 words from speeches given by the other Justice and find that the average word length in the sample is 4.1 words. Assume that the standard deviation of word lengths in speeches written by the other Justice is the same as the standard deviation of word lengths in speeches written by Justice Sonia Sotomayor.Historians are interested in comparing the word lengths for speeches by Justice Sonia Sotomayor to speeches by another Supreme Court Justice. They take a random sample of 50 words from speeches given by the other Justice and find that the average word length in the sample is 4.1 words. Assume that the standard deviation of word lengths in speeches written by the other Justice is the same as the standard deviation of word lengths in speeches written by Justice Sonia Sotomayor. Part A: Are the conditions for the Central Limit Theorem met in this case? Explain. Edit * Insert Formats B / U X X A A Part B: If the distribution of the word lengths for the other Justice had the same mean and standard deviation as that of Justice Sonia Sotomayor, use the normal approximation to calculate the probability of observing a sample mean of 4.1 words or less in a random sample of 50 words. Round to three decimal places. Mean = Standard Deviation P(4.1 or less) = Part C: Based on your answer to Part B, do the data appear to provide evidence that the average word length for this other Justice is smaller than the average word length for Justice Sonia Sotomayor? Explain Edit * Insert Formats B / U X X A A