5. The scoring averages (points per game) of the startinglineup of a basketball team are as follows: 6.2, 11.4, 18.3,22.9 and 14.8. What is the standard deviation of theiraverages?a. 7.14b. 6.39c. 5.72d. 5.226. A street-level road traffic estimates has beenconducted at the intersection from Monday to Friday.The following data shows the result: 48, 52, 40, 55, 60.What is the average daily flow for the week?a. 45b. 48c. 51d. 547. In a classroom, there are several students having thesame first name: Richard, 4; Liza, 2; James, 3; andKathryn, 5. Which name may be associated to the modeof the class?a. Richardb. Lizac. Jamesd. Kathryn

5) given data X:6.2,11.4,18.3,22.9,14.8 Here n=5 Average=Xbar =(1/n)*(sum(Xi)) =(1/5)*(73.6)…

Answered: lineup of a basketball team are as…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In the 1992 presidential election, Alaska's 40 election districts averaged 2002 votes per district for President Clinton. The standard deviation was 589. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places. a. What is the distribution of X? X N b. Is 2002 a population mean or a sample mean? Select an answer ✓ c. Find the probability that a randomly selected district had fewer than 2058 votes for President Clinton. d. Find the probability that a randomly selected district had between 1998 and 2225 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
In the 1992 presidential election, Alaska's 40 election districts averaged 1824 votes per district for President Clinton. The standard deviation was 573. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.a. What is the distribution of X? X ~ N(,)b. Is 1824 a population mean or a sample mean? c. Find the probability that a randomly selected district had fewer than 1872 votes for President Clinton. d. Find the probability that a randomly selected district had between 1842 and 2091 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
The scoring averages (points per game) of the starting lineup of a basketball team are as follows: 6.2, 11.4, 18.3, 22.9 and 14.8. What is the standard deviation of their averages? a. 7.14 b. 6.39 c. 5.72 d 5.22
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?
In the 1992 presidential election, Alaska's 40 election districts averaged 1993 votes per district for President Clinton. The standard deviation was 584. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places. a. What is the distribution of X? X ~ N( b. Is 1993 a population mean or a sample mean? Select an answer v C. Find the probability that a randomly selected district had fewer than 2053 votes for President Clinton. d. Find the probability that a randomly selected district had between 2074 and 2208 votes for President Clinton. e. Find the 95th percentile for votes for President Clinton. Round your answer to the nearest whole number.
A vending machine dispenses coffee into a twelve-ounce cup. The amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.04 ounce. You can allow the cup to overfil 3% of the time. What amount should you set as the mean amount of coffee to be dispensed? Cik to View ongn 1.o the table Click to view. Dagn 2.of the tabin. Dounces (Round to two decimal places as needed.)
the following numbers show the mileage of cars 10 sold . 86 , 87,90 , 87.2,88.2,87.4,87.8,89.7,88.9,89.6 find mean, standard deviation using probabilty plot. explain if data is normally distributed or not
In the 1992 presidential election, Alaska's 40 election districts averaged 2038 votes per district for President Clinton. The standard deviation was 576. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.a. What is the distribution of X? X ~ N(,)b. Is 2038 a population mean or a sample mean? Select an answer Population Mean Sample Mean c. Find the probability that a randomly selected district had fewer than 1842 votes for President Clinton. d. Find the probability that a randomly selected district had between 1997 and 2099 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
Please answer the following three multiple choice questions :) Summarized below are data for a study to see if hours of sleep per day for med and non-med students are the same: n x¯ s med students 25 5.8 0.83 non-med students 32 6.3 1.39 1. Find the test statistic (assume unequal standard deviation) A. -15.44 B. -5.69 C. -1.69 D. -1.81 2. Let tstat represent the value of the test statistic from the previous question. The degrees of freedom for the test are df = 52. To test the claim that the mean hours of sleep for med and non-med students is different, the P-value is calculated as: A. 2*pt( tstat, df=52, lower.tail = TRUE) B. 2*pt( tstat, df=52, lower.tail = FALSE) C. pt( tstat, df=52, lower.tail = TRUE) D. pt( tstat, df=52, lower.tail = FALSE) 3. The correct P-value is 0.098. At α = 0.05, we A. reject H0. The claim that means are different is not supported. B. reject H0. The claim that means are different is supported. C. fail to reject…
I need help with this question :) Here are the mean and Standard Deviation for both men and women: Men: Mean 69.0 Standard Deviation: 2.8 Women: Mean 63.6 Standard Deviation: 2.5 Tall Poppies International is a group for vertically gifted individuals (tall people). To join, a man must be at least74 inches tall, and a woman must be at least 70 inches tall.a. What percentage of men are eligible for membership?b. What percentage of women are eligible?c. Are the height requirements fair for men and women?
Consider the following data set. You may assume that the mean of this data set is x¯ = 7. 8 10 3 5 7 5 11 a) Find the range of the data set. b) Find the standard deviation of the data set. c) How many of the seven data points lie within one standard deviation of the mean?

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