Q2: According to the followingfrequency distribution table and raw:dataFrequency Distribution TableRaw DataClassf58| « • * *5669636271827990-99456775777263789480-8970-7911665067476074576660-6915537471585554807050-599.40-496460516168647266Q2 (d): According to the above rawdata, the value of mean deviation* :(MD) is8.0562 O8.1513 O8.1633 Onone of all above O

The mean deviation is the average of the absolute deviation of values from the mean for a given…

Answered: Q2 (d): According to the above raw…

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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A random sample of 1000 oranges showed that the mean amount of juice per orange was 8.2 fluid ounces, with a standard deviation of 1.5 fluid ounces. (a) Determine the z-score, to the nearest hundredth, of an orange that produced 6.8 fluid ounces of juice.(b) The z-score for one orange was 3.03. How much juice was produced by this orange? Round to the nearest tenth of a fluid ounce.
Suppose that the average height of men in the United States is approximately normallydistributed with a mean of approximately 70 inches and a standard deviation ofapproximately 4 inches. (d) What height is at the 20th percentile? (e) What height is at the 80th percentile? (f) What fraction of men’s heights are between the heights you found in (d) and (e)?
Please answer #14
Use the normal distribution of SAT critical reading scores for which the mean is 503 and the standard deviation is 113. Assume the variable x is normally distributed. (a) What percent of the SAT verbal scores are less than 650? (b) If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 575? (a) Approximately (Round to two decimal places as needed.) % of the SAT verbal scores are less than 650.
During the winter term, the same student's trip to their college takes 16.39 min by car, with a mean trip time for that term of 14.65 min and with a standard deviation of 4.07 min. What was their z-score for this particular winter trip? Round your answer to 2 decimal places
Evaluate the followings: (a) P (z<?)=0.0918 (b)Z.40 (c)P(1.44<z<2.33) (d) Z.99 (e) P (z>?) = 0.6950, (f) within 1.35 standard deviations from the mean (it means P (-1.35<Z<1.35)).
Per the Barron's report the average weeks unemployed is 21.5 (population mean) with a population standard deviation of 5 weeks and a sample of 55. In the next five questions we will be calculating the probability that the sample of 55 will provide a mean within 1 week of the population mean. What is the probability to 4 decimals for (x<20.5)? Use the rounded standard error of the mean in calculations.
The time taken by employees at Grace Floral shop to put a bouquet together has a normal distribution with mean 26.4 minutes and standard deviation .8 minutes. (i) Find the probability that an employee chosen at random takes between 24.6 and 27.8 minutes to put a bouquet together. [5] (ii) 12% of employees take more than t minutes to put a bouquet together. (ii) Find the value of t.
A randomly selected sample of n = 55 individuals over 65 years old takes a test of memorization skills. The sample mean is x = 53, and the standard deviation is s = 6.9. Give the numerical value of the standard error of the mean, s.e. (Round your answer to two decimal places.) s.e.(x) =
3

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Q2 (d): According to the above raw data, the value of mean deviation * :(MD) is 8.0562 O 8.1513 O 8.1633 none of all above O