In determining automobile mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 35 mpg and a standard deviation of 1.9 mpg. Complete parts a through e below.a. Find P(X < 34).P(X< 34) = |(Round to four decimal places as needed.)b. Find P(32 37).P(X> 37) = (Round to four decimal places as needed.)d. Find P(X> 34).P(X> 34) =|(Round to four decimal places as needed.)e. Find the mileage rating that the upper 11% of cars achieve.The mileage rating that the upper 11% of cars achieve ismpg.(Round to two decimal places as needed.)

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Answered: In determining automobile mileage…

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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Emissions of sulfur dioxide by industry sett off chemical changes in the atmosphere that result in "acid rain". The acidity of liquids is measured by pH on a scale of 0 to 14. Distilled water has pH 7.0, and lower pH values indicate acidity. Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below 5.0. Suppose that pH measurements of rainfall on different days in a Canadian forest follow a Normal distribution with standard deviation σσ = 0.5. A sample of n days finds that the mean pH is x⎯⎯⎯x¯ = 4.8. What are the P- Values for tests of sample sizes n = 5, n = 15 and n = 40? n = 5: n = 15: n = 40:
Please do the following question with program R.
The weight of a carton of a dozen eggs produced by a certain breed of hens is supposed to be normally distributed with a mean of 780 grams. A quality manager randomly checks thirty-five cartons of eggs (n = 35) to see whether the mean weight differs from 780 grams. She finds x̄ = 796. 1. What is the parameter of interest? The mean weight of 35 cartons of eggs produced by a certain breed of hens The mean weight of all cartons of eggs produced by a certain breed of hens The weight of all cartons of eggs produced by a certain breed of hens The mean weight of all eggs produced by a certain breed of hens 2.What are the null and alternative hypotheses? H 0: μ = 780 vs. H a: μ ≠ 780 H 0: μ = 780 vs. H a:μ < 780 H 0: μ = 780 vs. H a: μ > 780 H 0: x̄ = 780 vs. H a: x̄ ≠ 780
For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula is where P=1- 1/k2 p is the minimum percentage of scores that fall within kk standard deviations on both sides of the mean. Use this formula to answer the following questions. b) If you have scores that are normally distributed, find the percentage of scores that fall within 3.3 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 35 percent of the scores? Note: To answer part c you will need to solve the equation for k.k. A manufacturer knows that their items have a normally distributed length, with a mean of 5.3 inches, and standard deviation of 1.5 inches.If one item is chosen at random, what is the probability that it is less than 8.3 inches long?
Let X-G(6,0.3) Find the variance of the gamma distribution.
A variable of two populations has a mean of 55 and a standard deviation of 36 for one of the populations and a mean of 55 and a standard deviation of 10 for the other population. Complete parts (a) through (c). a. For independent samples of size 9 and 4, respectively, find the mean and standard deviation of x1−x2. (Assume that the sampling is done with replacement or that the population is large enough.) The mean of x1−x2 is _________ (Type an integer or a decimal. Do not round.) The standard deviation of x1−x2 is _________ (Round to four decimal places as needed.) b. Must the variable under consideration be normally distributed on each of the two populations for you to answer part (a)? Choose the correct answer below. A. No, the variable must be approximately normally distributed on at least one of the two populations for the formulas for the mean and standard deviation of x1−x2to hold, as long as the sampling is done with replacement. B. No, the variable…
Suppose that f (x) = (3/2)x2 for −1< x < 1. Determine the mean and standard deviation of X.
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 50 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that ? = 7.20 ml/kg for the distribution of blood plasma. (d) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.50 for the mean plasma volume in male firefighters. (Round up to the nearest whole number.) ______ male firefighters
For data that is not normally distributed we can't use Z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula 1 where p is the minimum percentage of scores that fall within k k2 is p = 1- standard deviations on both sides of the mean. Use this formula to answer the following questions. a) If you have scores and you don't know if they are normally distributed, find the minimum percentage of scores that fall within 3.3 standard deviations on both sides of the mean? b) If you have scores that are normally distributed, find the percentage of scores that fall within 3.3 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 98 percent of the scores? Note: To answer part c you will need to solve the equation for k.
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 44 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that ? = 7.30 ml/kg for the distribution of blood plasma.
Suppose the correlation, r, between two variables x and y is –0.44. What percentage of the variability in y cannot be explained by x?
Let x = (x1.X2 .. .x,. . ,,) be a data set with a sample mean x. Show that E(x, – x) = 0.

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