According to the data, the middle 60% of women will be between 62 and 66 inches tall. Explain how we would arrive at such a conclusion using normal distribution. You are encouraged to upload a picture to support your explanation.Heights of men and women in the U.S. are normally distributed. Recent information shows:Adult men heights: p= 69.6 inches with o = 3 inches.Adult women heights: p = 64.1 inches with o = 2.7 inches.Lab based on a lab pulled from: http://nc-stats.weebly.com/T T T F Paragı : Arial : 5 (18p :E- E- T--

The conclusion can be made by calculating the middle 0.6 probability that a selected women’s height.…

Answered: According to the data, the middle 60%…

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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If a data set has a normal distribution, what portion of the data falls between and 5+9 0.34 0.475 0.95 0.68
In the 1992 presidential election, Alaska's 40 election districts averaged 1946 votes per district for President Clinton. The standard deviation was 551. (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 1946 a population mean or a sample mean? c. Find the probability that a randomly selected district had fewer than 2004 votes for President Clinton. d. Find the probability that a randomly selected district had between 2029 and 2166 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
The figure is considered normal. The figure is considered positively skewed. The figure is considered negatively skewed. The figure is considered multimodal.
An instructor plots the grade distribution for a practice final exam and determines that the grades were negatively skewed. Based on this what would be your recommendation A. The entire class should have to attend additional practice tutoring sessions in order to bring up the average. B. Not enough information to make a recomendation, would need to know the average grade (mean). C. The mean was larger than the mode. D. The majority of the class did well on the exam.
Calculate a point estimate for the proportion of all MVCC students that have brown eyes. Do not round your answer.
The USDA prepared a report indicating that a family of four spent on an average $890 per month on food. Assume the distribution of food expenses for a family of four follows a normal distribution with a SD of $90 per month. What percent of the families spend more than $430 but less than $890 per month? What percent of the families spend between $830 and $1000 per month?
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A graphical display of a data set is given. Identify the overall shape of the distribution as (roughly) bell-shaped, triangular, uniform, reverse J-shaped, J-shaped, right skewed, left skewed, bimodal, or multimodal. A relative frequency histogram for the sale prices of homes sold in one city during 2006 is shown below. Relative Frequency 0.35 0.30 0.25 0.20 0.15 0.10 0.05- 0.00 10 15 20 25 30 35 Sale price of home ($10,000s) 40 45 O Left skewed O Reverse J-shaped O J-shaped O Right skewed «< Question 5 A Moving to another question will save this response. PgUp PoDn End PrtScn Home E9
The British Department of Transportation studied to see if people avoid driving on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations. The data for each location on the two different dates is in the table. Estimate the mean difference in traffic count between the 6th and the 13th using a 95% level. Round to the nearest whole number. Traffic Count Dates 1990, July 1990, July 6th 13th 124843 122766 122384 122909 1991, September 124255 122340 1991, September 123125 120366 1991, December 1991, December 127873 128871 118840 115788 135754 133695 1992, March 1992, March 131116 128582 1992, November 133042 130642 1992, November 126159 127110 With 95 count between the 6th and the 13th is between O % confidence, it can be said that the true mean difference in traffic and
In the 1992 presidential election, Alaska's 40 election districts averaged 1833 votes per district for President Clinton. The standard deviation was 558. (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(Correct,Correct) b. Is 1833 a population mean or a sample mean? Select an answer Sample Mean Population Mean c. Find the probability that a randomly selected district had fewer than 1915 votes for President Clinton. d. Find the probability that a randomly selected district had between 1813 and 1922 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
State whether you would expect the following distribution to exhibit low, moderate, or highvariation, and briefly explain: daily high temperatures in Philadelphia in March.

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