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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[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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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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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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AP/IB Biology Lab Assessment The effect of various fruit and vegetable cell membranes on their water potential Independent variable: Type of fruit or vegetable. (Produce used: Russet potatoes‚ Pascal celery‚ Gala apple‚ Navel orange‚ and Imperator carrot). The fruit or vegetable will be placed in six 56.7 gram cups‚ ranging with sucrose molarities of 0 (distilled water)‚ 0.2‚ 0.4‚ 0.6‚ 0.8‚ 1.0‚ with 5 trials‚ leading to 30 cups for each produce variable. Dependent variable: The water potential
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Current New Mean 270.275 267.5 Variance 76.61474291 97.94871868 Standard Deviation 8.7529848 9.8969045 Standard Error 1.383968415 1.564838 Using the formula‚ the interval estimation was calculated. Therefore‚ with confidence at 95% the differences between the mean distances are in between -1.385740214 and 6.935740214 yard. The sample size is said to have an inverse relationship with standard error: as the sample size increases‚ the
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manufacturing processes. In one particular application‚ a client gave Quality Associates a sample of 800 observations taken during a time in which that client’s process was operating satisfactorily. The sample standard deviation for these data was .21; hence‚ with so much data‚ the population standard deviation was assumed to be .21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples‚ the client could
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with a probability of over 99%; these plans are: CTC‚ HIC‚ and RCNC1. CTC costs are estimated to be the lowest when considering the average of $13.5m and standard deviation of only ~$2m which reflects a lower spread of the costs. Additionally‚ CTC cost savings average ~$14.5m when compared with the other plan offerings‚ with a low standard deviation which reflects more predictable savings. Similar results are obtained when we consider a safer accident rate (i.e.‚ 1 in 6‚6m flights). 2. Cost analysis
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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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the most appropriate descriptive statistics in the following table for the 3 variables of interest‚ as a starting point in the analysis: ASSETS YLD_7DY YLD_30DY Mean 1994.8089 4.1622 4.0982 Median 496.5000 4.1800 4.1300 Mode 1.70 4.16 3.89 Std. Deviation 4644.1251 .3262 .3253 Range 27003.90 2.12 2.12 Minimum 1.70 2.67 2.61 Maximum 27005.60 4.79 4.73 For assets‚ it is very clear that the range is very big; this is a first indication that the assets data has a lot of variability. Also‚ mean and median
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