A data set lists weights (lb) of plastic discarded by households. The highest weight is 5.67 lb, the mean of all of theweights is x = 2.242 lb, and the standard deviation of the weights is s = 2.067 lb.a. What is the difference between the weight of 5.67 lb and the mean of the weights?b. How many standard deviations is that [the difference found in part (a)]?c. Convert the weight of 5.67 lb to a z score.d. If we consider weights that convert to z scores between 2 and 2 to be neither significantly low nor significantly high,is the weight of 5.67 lb significant?a. The difference is lb.(Type an integer or a decimal. Do not round.)

The highest weight is 5.67 lb.Mean of all weights (x̄) = 2.242The standard deviation of weight (s) =…

Answered: A data set lists weights (lb) of…

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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4. From the data given below, find the number of items. Er = 90 r = 0.5 Exy = 120 Oy = 8 (where x and y are deviations from arithmetic average).
A scale is known to be "low-calibrated" in that it measures an aluminum bar to be 4 grams lower than the actual weight. If ????μold is the mean of all possible measurements of the bar using the old calibration, and ????μnew is the mean of all possible measurements of the bar using a re-calibrated scale. It is desired that ????−????=4μnew−μold=4. Consider the test ?0:????−????=4H0:μnew−μold=4 vs. ?0:????−????≠4H0:μnew−μold≠4 Sample after calibration: mean is 95.8 , SD is 4.9 and sample size is 10 Sample before calibration: mean is 89.352 , SD is 4.2 and sample size is 62 We do not know population standard deviations. What is the p-value? (Hint: best to use software, or else the Desmos Calculator for the 2 sample T and Z test)
A population of N=5 scores has a mean of μ=8. If one score with a value of X=4 is removed from the population, what is the value for the new mean? a. μ=9 b. μ=10 c. μ=8 d. μ=7
The average height of 1,000 children is 150cm with a standard deviation of 10cm. How many children stand between 155cm and 150cm ? A. Find the raw score for z= -1,-2,-3,+1,+2,+3 B. Sketch the graph C. What z score corresponds to 50? D. What z score corresponds to 60? E. Find the total area . F. Multiply the total area by 1000. G. Make a conclusion.
Suppose that 50 measurements were taken and the mean was calculated to be 4.55. Assume next that one of the data items was 19.4This was assumed to be an outlier as was discarded. What would be the mean of the remaining 49 numbers? I
A sample has a mean of M = 30. If one person with a score of X = 35 is removed from the sample, what effect will it have on the sample mean? Lütfen birini seçin: A. Cannot be determined from the information given B. The sample mean will decrease. C. The sample mean will increase. D. The sample mean will remain the same.
The data is in the height column. Answer 7 to 10
Excel's O a. COUNTIF function Ob. AVERAGE function O c. PivotTable report O d. SUM function can be used to construct a frequency distribution for quantitative data.
The authors of a paper were interested in how the distance a deer mouse will travel for food is related to the distance from the food to the nearest pile of debris. Distances were measured in meters. The data and computer output are given below. Distance from Debris Distance Traveled 6.94 0.00 5.23 6.13 5.21 11.29 7.10 14.35 8.16 12.03 5.50 22.72 9.19 20.11 9.05 26.16 9.36 30.65 Simple Linear Regression Results: Dependent Variable: Traveled Independent Variable: Debris Sample size: 9 R (correlation coefficient) = 0.5657 R-sq = 0.32002088 Estimate of error standard deviation 8.670711 Parameter estimates: Parameter Estimate Std. Err. Alternative DF T-Stat P-Value Intercept -7.6854587 13.332196 ≠ 0 7 -0.5764586 0.5824 Slope 3.2340908 1.7818117 ≠ 0 7 1.8150575 0.1124 a)What is the least squares regression line for the output given above? b) what is the predicted traveled distance given the distance from debris is 6.5 meters?
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Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.5 degrees. Low Temperature (o F) 40 - 44 45 - 49 50 - 54 55 - 59 60 – 64 D Frequency 2 9. ... The mean of the frequency distribution is degrees. (Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? O A. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. B. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is more than 5% of the actual mean. D. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean.

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