Combined test scores were normally distributed with mean 1492 and standard deviation 345. Find the combined scores that correspond to these percentiles.a) 35th percentile b) 75th percentile c) 80th percentileClick here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.Definitiona) The combined scores that correspond to 35th percentile is about(Round to the nearest integer as needed.)Entries in the table represent area under the curve between z = 0 and a positivecurve, area under the curve between z = 0 and a negative value of z are found inb) The combined scores that correspond to 75th percentile is about(Round to the nearest integer as needed.)Area = Probability0.000.010.020.030.04c) The combined scores that correspond to 80th percentile is about(Round to the nearest integer as needed.)= P(0 < x < 2)0.0.0000.0040.0080.0120.01600.1-12 dx.0398.0438.0478.0517.05570.2.0793.0832.0871.0910.0948Definition0.3.1179.1217.1255.1293.13310.4.1554.1591.1628.1664.17000.5.1915.1950.1985.2019.20540.6.2257.2291.2324.2357.2389Entries in the table represent area under the curve between z = 0 and a positive vacurve, area under the curve between z = 0 and a negative value of z are found in a0.7.2580.2611.2642.2673.27040.8.2881.2910.2939.2967.2995Area =Probability0.000.010.020.030.040.9.3159.3186.3212.3238.3264= P(0 < x < 2)1.5.4332.4345.4357.4370.4382.4452.4463.44741.0341334383461.3485.35081.6ex/2 dx1.7.4484.4495%3D.4554.4564.4573.4582.45911.1.3643.366536863708.37291.8.4641.4649.4656.4664.46711.2.3849.3869.3888.3907.39251.9.4713.4719.4726.4732.47381.3.4032.4049.4066.4082.40991.4.4192.4207.4222.4236.42512.0.4772.4778.4783.4788.47932.1.4821.4826.4830.4834.48382.2.4861.4864.4868.4871.48752.3.4893.4896.4898.4901.4904PrintDone2.4.4918.4920.4922.4925.4927

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Answered: Combined test scores were normally…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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The heights of Crepe Myrtle saplings at a local nursery are measured and the data recorded. The data is then plotted and the distribution is found to be bell-shaped. The mean of the set is 67 inches and the standard deviation is 3 inches. A graph of the situation is shown below: 58 61 64 67 70 73 76 a) What percentage of the saplings is more than 70 inches tall? b) What percentage of the saplings is between 61 and 73 inches tall?
Assume the selling prices of homes in a particular community follow a normal distribution with a mean of $263,100 and a standard deviation of $43,200. Complete parts a through c. a. Determine the range of selling prices in this community that includes one standard deviation around the mean. The range is from $ to $ (Type integers or decimals. Use ascending order.) b. Determine the range of selling prices in this community that includes two standard deviations around the mean. The range is from $ to $ (Type integers or decimals. Use ascending order.) c. Determine the range of selling prices in this community that includes three standard deviations around the mean. The range is from $ to $1. (Type integers or decimals. Use ascending order.)
The mean and median of a frequency distribution is 15 and 20 respectively. Then the mode of the distribution is O a. 17.5 O b. 16 O c. 20 O d. 30
A. Which one is greater the mean or the median? Which one is better measured of the center the median or the mean? Show work B. Find the mode C. Find the median D. Find the mean (even though we must round to the one decimal place to make computations easy round you’re mean to the nearest minute)
Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.10 to test for a difference between the weights of discarded paper (in pounds) and weights of discarded plastic (in pounds). Click the icon to view the data In this example, Hg is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis test? O A. H9: Ha =0 H1: Ha 0
❎❎❎❎✅✅✅✅✅✅✅
Use the Histogram tool to develop a frequency distribution and histogram for the number of months as a customer of the bank in the data below. Compute the relative and cumulative relative frequencies and use a line chart to construct an ogive. Complete the frequency distribution for the months as a customer, and compute the relative and cumulative relative frequencies. (Type integers or decimals rounded to three decimal places as needed.) Bins Frequency Relative Frequency Cumulative Relative Frequency 10 enter your response here enter your response here enter your response here 20 enter your response here enter your response here enter your response here 30 enter your response here enter your response here enter your response here 40 enter your response here enter your response here enter your response here 50 enter your response here enter your response here enter your…
Ch 8 FRQ ( The distribution of average speeds of the top 53 drivers in NASCAR is roughly symmetric, unimodal, and bell-shaped, with a mean of 184.8 mph and a standard deviation of 7.2 mph. In the space above, draw a Normal Curve, from z=-3 to z=+3. Draw vertical lines at the mean and each whole number standard deviation above and below the mean. Write two rows of numbers under the base line; one row with the number of standard deviations above and below the mean (z-scores), and a row under that with the actual values for this problem. ] What percent of drivers would you expect to have speeds below 199.2 mph ? Show all work. s] What is the area under the curve to the left of a speed of 170.4mph Show all work. What is the probability that a driver selected at random will have a speed below 172.0 mph ? Show all work. B] What percent of drivers would you expect to have a speed above 177.0 mph ? Show all work. How many drivers would you expect to have speeds between 170.4 mph and 177.0 mph…
The table available below shows the weights (in pounds) for a sample of vacuum cleaners. The weights are classified according to vacuum cleaner type. At a = 0.10, can you conclude that at least one mean vacuum cleaner weight is different from the others? Click the icon to view the vacuum cleaner weight data. Let μBU, BLU, and μTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? A. Ho: At least one of the means is different. Ha: MBUHBLUPTC B. Ho: "BU HBLU HTC # Ha: MBUHBLUPTC C. Ho: MBUHBLU = HTC Ha: At least one of the means is different. D. Ho: MBU - HBLU = μTC Ha: HBU #BLU HTC What is the test statistic? F = 12.56 (Round to two decimal places as needed.) What is the P-value? P-value = 0.000624 (Round to three decimal places as needed.)
I vacuum cleaners. The weights are classified according to vacuum cleaner type. At a=0.10, can you conclude that at least one mean vacuum cleaner weight is different The table available below shows the weights (in pounds) for a sample from the others? E Click the icon to view the vacuum cleaner weight data. Let Hgu. PBLU and PTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? O A. Ho: Heu * PBLU HTC H PBU PBLU =HTC Weight of Vacuum Cleaners by Type O B. Ho: At least one of the means is different. H PBu = HBLUu =PTC &c. Họ: PBU = HBLU =HTC Bagged upright 22 19 Bagless upright Top canister 21 22 23 25 H: At least one of the means different. 17 20 15 18 18 19 O D. Ho: PBU = PBLU "HTC 23 20 26 25 22| 23 Ha: Peu * PBLU *HTC What is the test statistic? Print Done F= (Round to two decimal places as needed.)
answer letter B using the above informations
The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates. Use the frequency distribution to approximate the population mean and population standard deviation. Compare these results to the actual mean u = $1.506 and standard deviation o = $1.004. Click the icon to view the frequency distribution for the tax rates. Frequency distribution of cigarette tax rates Tax Rate Frequency 0.00–0.49 7 0.50-0.99 14 1.00–1.49 7 1.50–1.99 2.00–2.49 5 2.50–2.99 5 3.00–3.49 3.50–3.99 1 4.00-4.49 1

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