The Z-score tells you the number of standard deviations away from the mean (and in whatX-μcan be used to find the Z-score for adirection) a data value is. The formula: Zsingle member of the population.Notice X - μ tells you the signed distance the data value (X) is from the mean. When youdivide that by the size of each chunk (the standard deviation) you are measuring thedistance in units of the size of the standard deviations (how many standard deviations fit inthe distance between the data value and the mean) - which gives you the Z-score.The formula X = μ+Z-o can be used to find the value in the population when giventhe Z-score (signed number of standard deviations).Notice that Z. tells you how far from the mean the data value is (since Z is the numberof standard deviations and o is the size of each standard deviation the product tells thetotal distance). So adding the distance from the mean to the mean gives the location alongthe number line for the data value.A certain population is normally distributed with a mean of 115 and a standard deviationof 12.How many standard deviations from the mean is a person in the population with a value of89. (That is find the z-score).

Given datamean=115standard deviation=12x=89

Answered: The Z-score tells you the number 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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The Z-score tells you the number of standard deviations away from the mean (and X-μ can be used to find the in what direction) a data value is. The formula: Z Z-score for a single member of the population. Notice X - μ tells you the signed distance the data value (X) is from the mean. When you divide that by the size of each chunk (the standard deviation) you are measuring the distance in units of the size of the standard deviations (how many standard deviations fit in the distance between the data value and the mean) - which gives you the Z-score. σ The formula X = μ+Z. can be used to find the value in the population when given the Z-score (signed number of standard deviations). Notice that Z. tells you how far from the mean the data value is (since Z is the number of standard deviations and is the size of each standard deviation the product tells the total distance). So adding the distance from the mean to the mean gives the location along the number line for the data value. A…
The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates. Use the frequency distribution to approximate the population mean and population standard deviation. Compare these results to the actual mean u = $1.481 and standard deviation a = $0.971. E Click the icon to view the frequency distribution for the tax rates. The population mean is S. (Round to three decimal places as needed.) Frequency distribution cigarette tax rates - X The population standard deviation is S (Round to three decimal places as needed.) Tax Rate 0.00-0.49 0.50-0.99 1.00-1.49 1.50-1.99 2.00-2.49 2.50-2. 99 3.00-3.49 3.50-399 4.00 4.49 Frequency O Compare these results to the values found using the actual data. 14 O A. The grouped values are both slightly larger. O B. The grouped values are both slightly' smaller. O C. The grouped mean is slightly larger, while the grouped standard deviation is slightly smaller. 17 O D. The grouped mean is slightly smaller, while the…
Find the z-score that has 68.4% of the distribution's area to its right. Can you also give the steps in the calculator? TI-84
Please help me solve this? Thanks
prove that If we multiply the data values by a constant (say, c), value of geometric mean will be affected. how this will affect the geometric mean? Use the equation in image to prove it, and give an example.
The coefficient of variation CV describes the standard deviation as a percent of the mean. Because it has no units, you can use the coefficient of variation to compare data with different units. Find the coefficient of variation for each sample data set. What can you conclude? Standard deviation CV = • 100% Mean E Click the icon to view the data sets. Data Table C/heights = 5.0 % (Round to the nearest tenth as needed.) Heights Weights CVweights =% (Round to the nearest tenth as needed.) 68 181 77 169 72 167 72 207 71 177 70 175 78 185 69 186 67 221 70 216 67 209 73 206
the mean = 146, and s = 35 Question #4 A score of 181 is how many z-scores above the mean?
In California, we need more rain to sustain the health of our natural environment, argriculture, and economic. A group of statistics students in Oxnard College recorded the amount of rain during 2016-2017 school year, measuring the intensity by the inches of rain, and the results were: Inches of Rain 1 2 3 4 5 6 Frequency 1 2 8 2 1 1 The mean (¯xx¯) rain intensity: inches (Please show your answer to 1 decimal place.)The median rain intensity: inchesThe mode rain intensity: inches (Please separate your answers by ',' in the bimodal situation. Enter DNE if there is no mode or if there are more than two modes.)
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.
IQ tests supposedly measure our intelligence. They typically are ‘normed’ so that they have a mean of 100 and a standard deviation of 15. (Higher scores show higher intelligence.) Use the graph and use the z-score and percents from columns b and c in Table A, when relevant. If you scored 132, what is your z-score? (Interpret.) Use table A X, z(a), (b), and (c) X z (a) (b) (c) and use the graph! Write down or save your response to use in Part 2 of this quiz. Group of answer choices -2.53 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. 2.13 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. -2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 0.57 percentile rank and 99.43% of people scored smarter than me. 2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 99.43 percentile rank…
The Z-score tells you the number of standard deviations away from the mean (and in what direction) a data value is. The formula: Z = *- can be used to find the Z-score for a single member of the population. Notice X - μ tells you the signed distance the data value (X) is from the mean. When you divide that by the size of each chunk (the standard deviation) you are measuring the distance in units of the size of the standard deviations (how many standard deviations fit in the distance between the data value and the mean) - which gives you the Z-score. The formula X = μ+Z.o can be used to find the value in the population when given the Z-score (signed number of standard deviations). Notice that Z. tells you how far from the mean the data value is (since Z is the number of standard deviations and is the size of each standard deviation the product tells the total distance). So adding the distance from the mean to the mean gives the location along the number line for the data value. A…

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