1h4.A fashion designer wants to know how many new dresses women buy each year. Asample of 650 women was taken to study their purchasing habits. Construct the 95%confidence interval for the mean number of dresses purchased each year if the samplemean was found to be 5.6. Assume that the population standard deviation is 1.3.Round your answers to one decimal place.Answer: Lower endpoint:Upper endpoint:inyir

From the provided information, Sample size (n) = 650 Population standard deviation (σ) = 1.3 Sample…

Answered: A fashion designer wants to know how…

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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On a nationwide math test, the mean was 57 and the standard deviation was 4. If Roberto scored 90, what was his z-score?
Please answer B, C, & D.
Suppose that the mean weight for men 18 to 24 years old is 170 pounds, and the standard deviation is 20 pounds. In each part, find the value of the standardized score (z-score) for the given weight. USE SALT (a) 206 pounds. Z = (b) 140 pounds. Z = (c) 173 pounds. Z = (d) 236 pounds. Z =
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 77.7 Mbps. The complete list of 50 data speeds has a mean of x = 18.55 Mbps and a standard deviation of s=31.38 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between 2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. The difference is Mbps. (Type an integer or a decimal. Do not round.)
There are two major tests of readiness for college, the ACT and the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT scores is approximately Normal, with mean of 21.2 and standard deviation of 5.2, SAT scores are reported on a scale from 400 to 1600. The distribution of SAT scores is approximately Normal, with mean of 1,159 and standard deviation of 205. What ACT scores make up the top 10% of all the scores? Round to the nearest whole number. 80 888 #3 $ & 3 4. 7 E Y U F G H J K DI
Cherise is interested in how her friends compare to the population for number of hours of television watched during a single week. She collects the data from 9 of her friends. The average for her friends is 28 hours, with a standard deviation of 9. She knows that the population mean is 34. What is the t statistic?
Test scores for a history class had a mean of 69 with a standard deviation of 4.5. Test scores for a physics class had a mean of 79 with a standard deviation of 3.5. Suppose a student gets a 75 on the history test and 83 on the physics test. Calculate the z- score for each test. On which test did the student perform better? A. B. C. D. History exam; history z-score is 1.33; physics z-score is 1.14 Physics exam; history z-score is 1.33; physics z-score is 1.14 History exam; history z-score is -1.33; physics z-score is -1.14 Physics exam; history z-score is -1.33; physics z-score is -1.14
Look for a,b,c and d. Thanks
you take a sample in equal 25 of test scores from a population. The sample mean is 38, and the population standard deviation is six. Five. What is the 95% confidence interval of the mean?
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 34 and the age of the winner for best actress was 43. For all best actors, the mean age is 42.5 years and the standard deviation is 6.2 years. For all best actresses, the mean age is 31.9 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Question content area bottom Part 1 Since the z score for the actor is z=enter your response here and the z score for the actress is z=enter your response here, the ▼ actress actor had the more extreme age. (Round to two decimal places.)
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 33 and the age of the winner for best actress was 47. For all best actors, the mean age is 44.4 years and the standard deviation is 6.5 years. For all best actresses, the mean age is 32.8 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z= and the z score for the actress is z = the V had the more extreme age. (Round to two decimal places.)
You are told that a set of data includes one score which is 0.8.The standard deviation for the set is 0. Can you give the mean of the set and say anything else about the six other scores in the set?

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