A national survey of 1600 adult citizens of a nation found that 22% dreaded Valentine’s Day. The margin of error for the survey was 1.7 percentage points with 90% confidence. Explain what this means.Which statement below is the best explanation? Select all that apply.- A. There is a 90% probability that the proportion of the adult citizens of the nation that dreaded Valentine’s Day is between 0.763 and 0.797.- B. There is 90% confidence that the proportion of the adult citizens of the nation that did not dread Valentine’s Day is between 0.763 and 0.797.- C. In 90% of samples of adult citizens of the nation, the proportion that dreaded Valentine’s Day is between 0.203 and 0.237.- D. There is 90% confidence that 22% of the adult citizens of the nation dreaded Valentine’s Day.- E. There is 90% confidence that the proportion of the adult citizens of the nation that dreaded Valentine’s Day is between 0.203 and 0.237.- F. There is 88.3% to 91.7% confidence that 22% of the adult citizens of the nation dreaded Valentine’s Day.

Answered: Image /qna-images/answer/6c50f4d0-8cf7-4d91-b048-f0e7177bf19b.jpg

Answered: A national survey of 1600 adult…

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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c. What is the likelihood of selecting a sample with a mean of at least $139,000? Note: Round your value to 2 decimal places and final answer to 4 decimal places. Probability d. What is the likelihood of selecting a sample with a mean of more than $127,000? Note: Round your value to 2 decimal places and final answer to 4 decimal places. Probability e. Find the likelihood of selecting a sample with a mean of more than $127,000 but less than $139,000. Note: Round your z-value to 2 decimal places and final answer to 4 decimal places
In the 1992 presidential election, Alaska's 40 election districts averaged 2118 votes per district for President Clinton. The standard deviation was 577. (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.1. Find the probability that a randomly selected district had between 2236 and 2434 votes for President Clinton. 2. Find the first 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 1823 votes per district for President Clinton. The standard deviation was 591. (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. c. Find the probability that a randomly selected district had fewer than 1803 votes for President Clinton. d. Find the probability that a randomly selected district had between 1930 and 2120 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
Answer the following questions accordingly. Take your time. Thanks in advance!
According to CTIA, 41% of all U.S. households are wireless-only households (no landline). Choose four households at random. A. What is the probability that at least two will be wireless-only? B. What is the probability that none of them will be wireless-only? C. What is the probability that least 1 of them will be wireless-only? D. Find the mean and standard deviation for the number of households that are wireless-only.
In the 1992 presidential election, Alaska's 40 election districts averaged 2008 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. Find the probability that a randomly selected district had fewer than 2096 votes for President Clinton. b. Find the probability that a randomly selected district had between 2047 and 2177 votes for President Clinton.c. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
A study by Correlated, Inc of 300 people found that about 51% of the sample liked brussels sprouts. Let p̂ be the proportion of people who like brussels sprouts. a. What is the mean of the sampling distribution of p̂? b. Find the standard deviation of the sampling distribution of p̂. Round to the nearest thousandth. c. Find the probability of selecting a sample of 300 people in which more than 55% of people like brussels sprouts.
Select the correct choice below and fill in the answer boxes to complete your choice (Type integers or decimals. Do not round.) A. One can be % confident that the mean drive-through service time of Taco Bell is between seconds and seconds. B. The mean drive-through service time of Taco Bell is of the time. OC. There is a seconds OD. One can be Taco Bell is % % probability that the mean drive-through service time of Taco Bell is between seconds and seconds % confident that the mean drive-through service time of seconds
Among employed women, 25% have never been married. Select 10 employed women at random. A. The number in your sample who have never been married has a binomial distribution. What is n?n? What is p?p? B. What is the probability that exactly 2 of the 10 women in your sample have never been married? C. What is the probability that 2 or fewer have never been married? D. What is the mean number of women who have never been married? E. What is the standard deviation?
According to Reader's Digest, 37% of primary care doctors think their patients receive unnecessary medical care. Use the z-table. a. Suppose a sample of 390 primary care doctors was taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. E(p) .37 (to 2 decimals) .0227 (to 4 decimals) b. What is the probability that the sample proportion will be within ±0.03 of the population proportion? Round your answer to four decimals. c. What is the probability that the sample proportion will be within ±0.05 of the population proportion? Round your answer to four decimals. d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? The probabilities would - Select your answer - This is because the increase in the sample size makes the standard error, o Select your answer -
The Transport Network Vehicle Service (TNVS) drivers have an average booking of 22 rides a day with standard deviation of 15 rides. If 100 TNVS will be randomly selected, find the average bookings which fall on the right side of the 75th percentile of the distribution. A 25 B 67 C) 75 D 23
Seventy-five percent of the students applying to a university are accepted. Using the binomial probability tables, what is the probability that among the next 18 applicants d. Fifteen or more will be accepted? e. Determine the expected number of acceptances. f. Compute the standard deviation.

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