Range = Maximum – minimum Interquartile range (IQR) = Q3 –Q1 Standard deviation is the average distance values fall from the mean of graph. 3. Q1(lower quartile) is the 25th percentile of ordered data or median of lower half of ordered data Median (Q2) is the 50th percentile of ordered data Q3 (upper quartile) is the 75th percentile of ordered data or median of upper half of ordered data Z scores are standarized standard deviation measurments of how far from the center (mean) a data value falls
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02+16.04+16.05+16.01+16.02+16.02+16.03+16.01+16+16.07)/12=16.03 (b) Calculate the sample standard deviation. =0.0193 3.2. The bore diameters of eight randomly selected bearings are shown here (in mm): 50.001‚ 50.002‚ 49.998‚ 50.006‚ 50.005‚ 49.996‚ 50.003‚ 50.004 (a) Calculate the sample average. (50.001+50.002+ 49.998+50.006+50.005+49.996+50.003+50.004)/8=50.001 (b) Calculate the sample standard deviation. =3.2186 3.4. Consider the furnace temperature data in Exercise 3.3 (The nine measurements
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[ασ (RA ) − (1 − α)σ (RB )]2 . Therefore‚ the standard deviation of portfolio P is σ (RP ) = ασ (RA ) − (1 − α)σ (RB ). As assets A and B are perfectly negatively correlated‚ we can construct portfolio P such that its standard deviation is 0. The weights of such portfolio are 0 = ασ (RA ) − (1 − α)σ (RB ) = 0.14 × α − 0.23 × (1 − α). Solving the above equation for α gives α= 0.23 = 0.622. 0.14 + 0.23 Portfolio P with standard deviation zero has weight 0.622 on asset A and weight 0
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the mean parts per million of dissolved solids in a 25-week sample to be 52 with a standard deviation of 32. Has there been a change in the average level of dissolved solids in this stream? 2 An exhaustive survey of all users of a wilderness park taken in 1960 revealed that the average number of persons per party was 2.6 In a random sample of 25 parties in 1985‚ the average was 3.2 persons with a standard deviation of 1.08. a. Test the hypothesis that the number of persons per party has changed
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length of labor in this study descriptive‚ this being represented by mean and standard deviation in the data. Yes‚these are appropriate as they both can be calculated (varible n=30 and mean=14.63) at the interval level of measurements. 3) Nominal level data could be used to describe the length of labor. This would have done by using frequencies‚ percentages‚ and mode instead of using mean‚ range‚ and standard deviation 4) No‚ the distributions of scores were not similar for the two groups. Experimental
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3.42 | Standard Error | 0.24593014 | Median | 3 | Mode | 2 | Standard Deviation | 1.73898868 | Sample Variance | 3.02408163 | Kurtosis | -0.7228086 | Skewness | 0.52789598 | Range | 6 | Minimum | 1 | Maximum | 7 | Sum | 171 | Count | 50 | Frequency Distribution: | Size | Frequency | 1 | 5 | 2 | 15 | 3 | 8 | 4 | 9 | 5 | 5 | 6 | 5 | 7 | 3 | The median of the data is 3 and the mode is 2. The household size mean is 3.42. The standard deviation is 1.74 and
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We can do a lot of good statistics with the normal curve‚ but virtually none with any other curve. Let us assume that we have recorded the 1000 ages and computed the mean and standard deviation of these ages. Assuming the mean age came out as 40 years and the standard deviation as
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quartiles; variance; standard deviation. Solution: A sample is a subset of a population. A population consists of every member of a particular group of interest. The variance and the standard deviation require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old children in Mrs. Jones ’ 2nd grade class. {25‚ 60‚ 51‚ 47‚ 49‚ 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. Solution: mean = 46
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1) Unstructured interviews are best used for D. asking individualized questions with no specific order 2) A recent study of breast cancer revealed that 13% of women in the sample used antibiotics more than 500 days in their lifetime. Further‚ 79% of these women developed breast cancer. According to the American Cancer Society‚ one in 12 women will develop breast cancer at some time in their lives. Of the numbers mentioned‚ which are parameters? D. 79% and one in 12 3) Basic business research
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of the sample| d.|mean of the population| ANS: A 5. A simple random sample of 100 observations was taken from a large population. The sample mean and the standard deviation were determined to be 80 and 12 respectively. The standard error of the mean is a.|1.20| b.|0.12| c.|8.00| d.|0.80| ANS: A 6. A population has a standard deviation of 16. If a sample of size 64 is selected from this population‚ what is the probability that the sample mean will be within 2 of the population mean? a.|0.6826|
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