1)From Nov.20th-Dec.31, my portfolio loss 5.41%. I can tell that I am losing 5.41% of my money that I invested and that I did not gain any money at all. Calculation: $100,000.00-$94,585.40=5,414.60/100,000.00=0.0541x100%=5.41% 2)Affiliated Group had the largest standard deviation, while Adobe had the smallest standard deviation. Affiliated has higher SD which mean that it has more variability and also higher risk. While Adobe has smallest SD which mean lower variability and lower risk of losing money. Both of the graphs started well at the beginning but then during the 2nd weeks of December, both dropped and then went up a little but then dropped again. Overall both graph seem similarly the same. There seem to be no outliers for both of the graphs. There are no outliers because the numbers variables seem to be regularly closed to each other and not far distance. …show more content…
There are no outliers for both of the box plots. According to the boxplot, Adobe the number seem to be symmetrical. The graph is comparatively short. While Affiliated, is screwed to the left. 4) My portfolio is losing more money when comparing to DJ Industrial Avg.,S&P 500 and the NASDAQ. My stocks in the NASDAQ is still losing money just like the other stocks in my portfolio. 5)Empirical Rule: For Adobe anything higher than 96.28 or lower than 89.10 is unusual. For Affiliated anything above 184.95 or below 145.71 is unusual. Adobe 92.69 +(1.793x2)=96.28 Affiliated 165.33+(9.809x2)=184.95 92.69-(1.793x2)=89.10 165.33-(9.809x2)=145.71 Z-Score: The Z-Score are not unusual because it is within the 2 standard
All the results show that the answers for variable Y have larger number than to variable X in 2012 and even larger compared to variable in 1990. Overall the one part that is remotely similar is the box and whisker plot; even though the numbers are not the same, the shapes of each of them are alike. Throughout all the measures, there is a noticeable pattern between the variables. Carbon emission has a larger result than the population per km2 in 2012 and 1990. The best measure to observe and best describes the central tendencies in both variables is the median, because the value is not distorted by the major outliers in the data set. Plus, since there are equal amount of data and both variables have a min lower than 5 and a max in the thousands, it seems fair that the median is the best central tendencies to describe the results; it is the only one that is snot
Question 6. The mean return for the Vanguard Total Stock Index is 20.8 while the mean return for the Vanguard Balanced Index is 12.9 (with bonds). Based on this data you would conclude that bonds do not reduce the overall risk of an investment portfolio since the mean return was actually less when the porfolio has bonds in
The following Probability Plot clearly exposes any outliers; data entries that are . . . “far removed from the other entries the data set”, (Larson & Farber, 2011. pg. 68). We can see from this plot that there are several outliers on the high and low end of the data. However, one outlier in particular could cause the conclusions made about the data to be flawed. The data point located at 98%, 4600, is of major concern, as it is causing this set of data to appear much higher. It is recommended that this entry be ignored when making conclusions about the data set as a whole.
1. In the spreadsheet Markowitz-01.xlsx some of the entries in the long/short and longonly portfolio data sections are missing (the missing data locations are highlighted in
Your net portfolio is approximately $1,330,000. This does not include what you reported to be approximately $250,000 in various taxable accounts or any checking, savings or other accounts you may have.
The standard Deviation comparing both sets of data only shows a slight difference on the numbers (a difference of only 1.38187054), but in the histograms both show a huge difference on the frequencies of each data set, but shows a similarity on the cumulative percentage. Data set #2 seem to be more stable even though the grades of the student were lower than data set #1. Data set#2 shows a relative stability in regards the frequency on the histograms, when comparing the frequency histograms data set #2 has more frequency on number two’s than data set #1 on frequency on the number three’s. In this case Data set #2 shows a normal distribution behavior.
She thought, “It’s obvious that the question of interest is the financial impact incurred relative to the benchmark of a well-managed portfolio. It was pretty clear that Martin’s portfolio had experienced significant losses. As to the data, the year-end statements are all that we need. The comparison will be easy”.
(a). Plot mean chart (x-chart) and range chart (R-chart) to analyze these data. Sample No. Observations Mean Range
The stock market simulation game is an online game that teaches students how to invest their money. The game gives each student 100,000 dollars to start investing in different stocks. It was very important to diversify the money into different stocks, because a person does not want to spend all their money in one place. The game is played over a six weeks period, and in that six weeks we lost 8,832.53 dollars. We may have lost money; however, we invest into twenty-eight different stocks.
The Standard Deviation is much smaller at 6.2% compared to Stock A. “Standard deviation is calculated based on the mean.” (WHAT IS STANDARD DEVIATION, n.d.) The space of each data point from the mean is squared, summed and averaged to find the variance. The difference is a result of taking the mean of the data points, subtracting the mean from each data point individually, squaring each of these results and then taking another mean of these squares. Ordinarily, the smaller a stock’s standard deviation, the less volatile or risky it is. The higher the standard differential, the more scattered the profits are, consequently is more dangerous. The additional 1% of the profit produced by Stock A is not worth the risk associated with Stock
(1) Calculate the expected return (rp), the standard deviation (p), and the coefficient of variation (CVp) for this portfolio and fill in the appropriate blanks in the table.
For the month of December, I was given an assignment consisting of $100,000 and four stocks to invest in. My four stocks were The Ralph Lauren Corp., Visa Inc., Master card Inc. and The Chevron Corp. As stated I was given a month to record my data and I ended up with a total capital gain of $5,518.36 for the one month period for my investments. I have to thank you Mr. Acker, this project was not difficult, but it did confuse me. Receiving this assignment scared me in a way, because I didn’t know what I was getting into. The finance world is scary and tricky, one minute the market is doing good and other days it would be low. While calculating my capital gains or losses I thought I would lose a larger
These represent the range of the sale price. Lastly, I used the formula to get the standard deviation 48,945.28, which measures the variability.
On the graph above you can see that both the quadratic and the line are both adequate representations of the data collected by gold medalists for the men’s high jumps in the Olympics. Both of these lines follow the plots made on the original graph and they don’t stray too far from those lines either. There only outlier for the quadratic seem to be the medalist from the 1948 Olympics because his height is far below the quadratic. There might be some problems with the exact position of where the quadratic is and where the line is because they were drawn by hand and not on the computer like the stat plots which could potentially cause problems for interpreting the data.