This website uses cookies to ensure you have the best experience.

Mean And Standard Deviation Essay

886 words - 4 pages

Mean and standard deviation

The median is known as a measure of location; that is, it tells us where the data are. As stated in , we do not need to know all the exact values to calculate the median; if we made the smallest value even smaller or the largest value even larger, it would not change the value of the median. Thus the median does not use all the information in the data and so it can be shown to be less efficient than the mean or average, which does use all values of the data. To calculate the mean we add up the observed values and divide by the number of them. The total of the values obtained in Table 1.1 was 22.5  , which was divided by their number, 15, to give a mean of 1.5. ...view middle of the document...

The Normal distribution is represented by a family of curves defined uniquely by two parameters, which are the mean and the standard deviation of the population. The curves are always symmetrically bell shaped, but the extent to which the bell is compressed or flattened out depends on the standard deviation of the population. However, the mere fact that a curve is bell shaped does not mean that it represents a Normal distribution, because other distributions may have a similar sort of shape.
Many biological characteristics conform to a Normal distribution closely enough for it to be commonly used - for example, heights of adult men and women, blood pressures in a healthy population, random errors in many types of laboratory measurements and biochemical data. Figure 2.1 shows a Normal curve calculated from the diastolic blood pressures of 500 men, mean 82 mmHg, standard deviation 10 mmHg. The ranges representing [+-1SD, +12SD, and +-3SD] about the mean are marked. A more extensive set of values is given in Table A of the print edition.
Figure 2.1

The reason why the standard deviation is such a useful measure of the scatter of the observations is this: if the observations follow a Normal distribution, a range covered by one standard deviation above the mean and one standard deviation below it

includes about 68% of the observations; a range of two standard deviations above and two below () about 95% of the observations; and of three standard deviations above and three...

Other Papers Like Mean and Standard Deviation

Goniometry Essay

1057 words - 5 pages SHOULDER FLEXION Test Position * Subject supine * Flatten lumbar spine (flex knees) * Shoulder no abduction, adduction or rotation * (note: to measure gleno-humeral motion, stabilize scapula)   | Normal Range(for shoulder complex flexion) * 167o  + or -  4.7o (American Academy of Orthopaedic Surgeons) * 150o (American Medical Association) * 166o (mean), 4.7o (standard deviation), (Boone and Azen) | Goniometer Alignment

Week 2 Practice Problems Essay

1643 words - 7 pages ------------------------------------------------- University of Phoenix Material Week Two Practice Problems Prepare a written response to the following questions. Chapter 2 12. For the following scores, find the mean, median, sum of squared deviations, variance, and standard deviation: 1,112; 1,245; 1,361; 1,372; 1,472 Mean: 1,312.4 Median: 1,361 Sum of squared deviation: 76,089.2 Variance: 15,217.84

Bba307 5.1

620 words - 3 pages 1. For each company A. Deterring the mean and standard deviation of the returns. B. Calculate the coefficient of variation. C. Determine which company appears to be more volatile with respect to its risk. D. Identify the company with which you would choose to invest. Part a: Year | Company A Return | Company B Return | Average Market Return | Company A Deviation | Company A Deviation Squared | Company B Deviation | Company

Practice Quiz Week 7 Statistics

1999 words - 8 pages Math 221 Quiz Review for Weeks 5 and 6 1. Find the area under the standard normal curve between z = 1.6 and z = 2.6. 2. A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was \$61,400 with a standard deviation of \$2,200. Find a 95% confidence interval for the true mean annual income of the business’ customers. 3. IQ test scores are normally

Runereavers

668 words - 3 pages University of Phoenix Material Time to Practice – Week Two Complete Parts A, B, and C below. Part A 1. Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions? 2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98. |Raw score |Z score | |68.0 |? | |? |–1.6

Statistic

1112 words - 5 pages variable is discrete. Also we only have a 1/6 chances when rolling a number. | 2. Calculate the mean and standard deviation of the probability distribution created by rolling a die. Either show work or explain how your answer was calculated. Mean: 3.5 Standard deviation: 1.707Mean = 1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6) = 3.5Standard Deviation= (1-3.5)^2(1/6)+(2-3.5)^2(1/6)+(3-3.5)^2(1/6)+(4-3.5)^2(1/6)+(5-3.5)^2(1/6)+(6-3.5)^2(1

Case Problem Quality Associates, Inc

679 words - 3 pages determine what action, if any, should be taken. Provide the test statistic and p-value for each test. 2. Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable? 3. Compute limits for the sample mean x ̅ around μ = 12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If x ̅ exceeds the

Htl-362v Homework

807 words - 4 pages HLT362V Week 1 Homework EX#16 Answers for EXERCISE 16 page 122 (Questions 1- 4 are optional)• Mean and Standard Deviation Exercise 16: Mean and Standard Deviation 1. The researchers analyzed the data they collected as though it were at what level of measurement? a. Nominal b. Ordinal c. Interval/ratio d. Experimental Answer: c. The researchers analyzed the data as though it were at the interval/ratio level since they calculated means

Determination of Lead and Zinc Using Aas with Standard Additions

823 words - 4 pages Concentration of Zinc (ppm) | Absorbance | 0 | -0.004 | 0.5 | 0.2873 | 1 | 0.5031 | 1.5 | 0.6427 | 2 | 0.7519 | 2.5 | 0.8247 | Figure 2: Absorbance readings of unknown Zinc concentration and their mean, standard deviation, and standard error of their mean Unknkown Absorbances (ppm) | Mean unknown absorbance (ppm) | Standard Deviation | Standard Error of Mean | 0.7992 | 0.801567 | 0.002122 | 0.001225 | 0.8033

Fi 4000 Case Report

546 words - 3 pages , I calculated monthly returns and standard deviation of monthly returns for each stock. After I got my monthly returns and standard deviation I converted them to annual values by using formulas. Formulas that were given are below: Annualized Mean Return= (1+Monthly Mean Return) ^12-1 Annualized Standard Deviation= Monthly Standard Deviation *square root of (12) Calculation results are presented below: 2b)By using my annual returns and

Wk Qnt561

699 words - 3 pages is 67 – 23 = 44 Standard deviation = sqrt([(28 - 40.84)^2 + (39 - 40.84)^2 + (...same pattern) + (29 - 40.84)^2] / 25 ) =14.55 82. a. Mean = (3 x 90 + 8 x 110 + 12 x 130 + 16 x 150 + 7 x 170 + 4 x 190) / 50 = 141.2 b. Standard deviation =sqrt((3(90 - 141.2)^2 + 8(110 - 141.2)^2+ (...same pattern) +4(190 141.2)^2) / 50) =26 c. Within two standard deviations  95% 141.2 – 2 x 26.2, 141.2 + 2 x 26.2 Limits between 88.8 and 193.6 87. I. a. Mean

Related Essays

Statistics Math 221 Lab Week 4

937 words - 4 pages class survey results that were entered into the MINITAB worksheet. 2. Calculate descriptive statistics for the variable where students flipped a coin 10 times. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to the coin. The output will show up in your Session Window. Type the mean and the standard deviation here. Mean: 4.6Standard deviation: 1.429 | Short Answer Writing Assignment – Both the

Standard Deviation Essay

1640 words - 7 pages distribution in a completely different way. Idea The idea behind the standard deviation is to quantify the spread of a distribution by measuring how far the observations are from their mean, x. The standard deviation gives the average (or typical distance) between a data point and the mean, x. Notation There are many notations for the standard deviation: SD, s, Sd, StDev. Here, we'll use SD as an abbreviation for standard deviation, and

Walt Disney Bonds Essay

1314 words - 6 pages Assignment #1: Virginia Capital Portfolio Optimization Stock | Mean of Return | Standard Deviation | IBM | 1.27% | 9.38% | 1) The below table shows the mean of IBM’s Stock to be 1.27% and the standard deviation to be 9.38%. Stock | Mean | Standard Deviation | Complete Portfolio of Weighted S&P, Lehman, and MSCI | 0.67% | 2.75% | 2) The table above shows the mean and standard deviation of a portfolio with S&P 500

Disertation Essay

2911 words - 12 pages understand the evolution process and to use empirical evidence of natural selection to find out how new species emerged (Magnello, 2005). USES OF STANDARD DEVIATION Standard deviation is defined as a measure of dispersion of a set of data from its mean (Investopedia, 2010). It is calculated as the square root of variance. The more spread apart the data, the higher the deviation. It is also seen as a measure of variability among the values of a frequency