Answered: A sample of 10 networking sites for a…

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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5. Letitia swam the girls 100m freestyle in 56.5 seconds. Ben swam the boys 100m freestyle in a time of 50.3 seconds. The mean time for the girls 100m at the swim meet was 58.6 seconds with a standard deviation of 2.3 seconds. The mean time for the boys 100m at the swim meet was 52.1 with a standard deviation of 1.8. Who swam the faster relative time? esc 180 1 27 @ 2 #3 13 $ OCT do % M ^
Suppose a normally distributed set of stock prices with 4100 observations has a mean of 114 and a standard deviation of 16. Use the 68-95-99.7 Rule to determine the number of observations in the data set expected to be between the values 66 and 146. Hint: This problem is asking for how many observations, not the percent. Answer = (Round to the nearest whole observation.)
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.
Total enrollment in the Kalamazoo Public Schools have a mean and standard deviation of 11,957 and 13, respectively. The district expects growth in the next five years, so every observation in the dataset is multiplied by 1.5. What will the new mean be? 1) 7,971.33 2) 1.5 3) 11,957 4) 17,935.5 5) 11,958.5
12
Office projects have a mean time of 114 days to be completed with a standard deviation of 456 hours. Around how many days they worked on a project if 41.66 percent of the project left to be completed? O A. 41.66 OB. 58.34 O C. 110 OD. 118
Answer part c and d frq styled
b) Rachel received the following scores on the midterm and final examinations in his introductory psychology course. The mean and the standard deviation for each test are given below. Assume that the normal distribution provides a good model for the frequency distribution. Section 8. 40 90 |Midterm 50 Final 100 12 i. What percentage of students scored better than Rachel on the midterm? The department has decided that all marks from introductory psychology courses should be reported based on a standardized distribution with a mean of 100 and standard deviation of 15. What would be Rachel's final examination scores based on ii. this new distribution?
If the mean speed for a major league fast ball is 90 miles per hour and the standard deviation is 5 miles per hour, what would the standardized score be for a pitch that is 97 miles per hour? O 0.4 4.0 1.4 O 1.0
Suppose the weight of dogs is normally distributed with a mean of 35 pounds and a standard deviation of 5 pounds. What is the z-score for a dog weighing 34 lbs? Group of answer choices -0.2 0.2 0.98 -0.98 None of the above
Consider a data set that has a mean of 82.7 and a standard deviation of 14.1. According to Chebyshev's Theorem, give an interval that has at least 91.14% of the data.
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…

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A sample of 10 networking sites for a specific month has a mean of 26.1 and a standard deviation of 4.2. Which of these is the point estimator? a. 4.2 b. 1.83 c. 3.25