Let R * T = 72, where R = the rate of growth (for example, interest rate), T = doubling time (for example, time it takes to double an amount of ).
Plug in the value for R = rate of growth. For example, how long does it take to double $100 to $200 at an interest rate of 5% per annum? Substituting R = 5, we get 5 * T = 72.
Solve for the . In the example given, divide both sides by R = 5, to get T = 72/5 = 14.4. So it takes 14.4 years to double $100 to $200 at an interest rate of 5% per annum.
Study these additional examples:
How long does it take to double a given amount of money at a rate of 10% per annum? Let 10 * T = 72, so T = 7.2 years.
How long does it take to turn $100 to $1600 at a rate of 7.2% per annum? Recognize that it takes 4 doubling to get from $100 to $1600 (double of $100 is $200, double of $200 is $400, double of $400 is $800, and double of $800 is $1600). For each doubling, 7.2 * T = 72, so T = 10.
that by 4 yields 40 years.
Let R * T = 72, where R = the rate of growth (for example, interest rate), T = doubling time (for example, the time it takes to double an amount of money).
Plug in value for T = doubling time. For example, if you want to double your money in ten years, what
do you need? Substituting T = 10, we get R * 10 = 72.
Solve for the unknown variable. In the example given, divide both sides by T = 10, to get R = 72/10 = 7.2. So you will need 7.2% annual interest rate to double your money in ten years.
Estimate the time to lose half of your capital: as in the case of inflation. Solve T = 72/R, after plugging in value for R, analogous to estimating doubling time for exponential growth (it's the same as the doubling formula, but you think of the result as inflation rather than growth), for example:
How long will it take for $100 to depreciate to $50 at an inflation rate of 5%?
Let 5 * T = 72, so 72/5 = T, so that T = 14.4 years for buying power to halve at an inflation rate of 5%.
Estimate the rate of decay for a certain time span: Solve R = 72/T, after plugging in value for T, analogous to estimating growth rate for exponential growth, for example:
If the buying power of $100 becomes worth only $50 in ten years, what is the inflation rate per annum?
Let R * 10 = 72, where T = 10 so that we may find R = 72/10 = 7.2% for that one example.
Beware! a general trend (or average) of
– and "out of bounds," outliers, or odd examples are simply ignored, and dropped out of consideration.
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Felix's Corollary to the Rule of 72 is used to approximate the future value of an annuity (a series of regular payments). It states that the future value of an annuity whose percentage interest rate and number of payments multiply to be 72 can be approximated by multiplying the sum of the payments times 1.5. For example, 12 periodic payments of $1000 growing at 6% per period will be worth approximately $18,000 after the last period. This is an application of Felix's Corollary to the Rule of 72 since 6 (the percentage interest rate) times 12 (the number of payments) equals 72, so the value of the annuity approximates 1.5 times 12 times $1000.
The value 72 is chosen as a convenient choice of numerator, since it has many small divisors: 1, 2, 3, 4, 6, 8, 9, and 12. It provides a good approximation for annual compounding, and for compounding at typical rates (from 6% to 10%). The approximations are less exact at higher interest rates.
Let the rule of 72 work for you, by starting saving now. At a growth rate of 8% per annum (the approximate rate of return in the stock market), you would double your money in 9 years (8 * 9 = 72), quadruple your money in 18 years, and have 16 times your money in 36 years.
For periodic compounding, FV = PV (1 + r)^T, where FV = future value, PV = present value, r = growth rate, T = time.
If money has doubled, FV = 2*PV, so 2PV = PV (1 + r)^T, or 2 = (1 + r)^T, assuming the present value is not zero.
Solve for T by taking the natural logs on both sides, and rearranging, to get T = ln(2) / ln(1 + r).
for ln(1 + r) around 0 is r - r2/2 + r3/3 - ... For low values of r, the contributions from the higher power terms are small, and the expression approximates r, so that t = ln(2) / r.
Note that ln(2) ~ 0.693, so that T ~ 0.693 / r (or T = 69.3 / R, expressing the interest rate as a percentage R from 0-100%), which is the rule of 69.3. Other numbers such as 69, 70, and 72 are used for easier calculations.
For periodic compounding with multiple compounding per year, the future value is given by FV = PV (1 + r/n)^nT, where FV = future value, PV = present value, r = growth rate, T = time, and n = number of compounding periods per year. For continuous compounding, n approaches infinity. Using the definition of e = lim (1 + 1/n)^n as n approaches infinity, the expression becomes FV = PV e^(rT).
If money has doubled, FV = 2*PV, so 2PV = PV e^(rT), or 2 = e^(rT), assuming the present value is not zero.
Solve for T by taking natural logs on both sides, and rearranging, to get T = ln(2)/r = 69.3/R (where R = 100r to express the growth rate as a percentage). This is the rule of 69.3.
For continuous compounding, 69.3 (or approximately 69) gives more accurate results, since ln(2) is approximately 69.3%, and R * T = ln(2), where R = growth (or decay) rate, T = the doubling (or halving) time, and ln(2) is the natural log of 2. 70 may also be used as an approximation for continuous or daily (which is close to continuous) compounding, for ease of calculation. These variations are known as rule of 69.3, rule of 69, or rule of 70.
A similar accuracy adjustment for the rule of 69.3 is used for high rates with daily compounding: T = (69.3 + R/3) / R.
The Eckart-McHale second order rule, or E-M rule, gives a multiplicative correction to the Rule of 69.3 or 70 (but not 72), for better accuracy for higher interest rate ranges. To compute the E-M approximation, multiply the Rule of 69.3 (or 70) result by 200/(200-R), i.e., T = (69.3/R) * (200/(200-R)). For example, if the interest rate is 18%, the Rule of 69.3 says t = 3.85 years. The E-M Rule multiplies this by 200/(200-18), giving a doubling time of 4.23 years, which better approximates the actual doubling time 4.19 years at this rate.
The third-order Padé approximant gives even better approximation, using the correction factor (600 + 4R) / (600 + R), i.e., T = (69.3/R) * ((600 + 4R) / (600 + R)). If the interest rate is 18%, the third-order Padé approximant gives T = 4.19 years.
To estimate doubling time for higher rates, adjust 72 by adding 1 for every 3 percentages greater than 8%. That is, T = [72 + (R - 8%)/3] / R. For example, if the interest rate is 32%, the time it takes to double a given amount of money is T = [72 + (32 - 8)/3] / 32 = 2.5 years. Note that 80 is used here instead of 72, which would have given 2.25 years for the doubling time.
Here is a table giving the number of years it takes to double any given amount of money at various interest rates, and comparing the approximation with various rules:
Don't let the rule of 72 work against you, when you take on high interest debt. Avoid credit card debt! At an average interest rate of 18%, the credit card debt doubles in just 4 years (18 * 4 = 72), and quadruples in only 8 years, and keeps escalating with time. Avoid credit card debt like the plague.
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Becomean Author!follow the rule是什么意思_百度知道
follow the rule是什么意思
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follow the rule工作规则；要守规矩；按规则办事例句1.She sensibly promises that she will follow the rule of law.她明智地承诺她会遵从法律。2.You should try to follow the rule of &the more, the faster, the better& when trying to lower cholesterol.降低胆固醇更应该把握「越多、越快、越好」的原则噢。3.Water does not a哗担糕杆蕹访革诗宫涧lways follow the rule that everything expands when heated and contracts when cooled.水并不总是遵循物质热胀冷缩的规律。4.Always follow the rule and guideline as directed by your volunteer coordinator.遵循规则、指导方针和志愿协调员的指挥。5.What's more, it requires us to follow the rule of history teaching through learning of enquiry.它还要求我们遵循研究性学习下的历史教学原则。6.Everything must follow the rule, in order and must be right.每件事都必须按规矩办，有秩序，并且是正确的。
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How to decide skewness by looking at a boxplot built from this data:
340, 300, 520, 340, 320, 290, 260, 330
One book says, "If the lower quartile is farther from the median than the upper quartile, then the distribution is negatively skewed." Several other sources said more or less the same.
I built a boxplot using R. It's like the following:
I take that it's negatively skewed, because the lower quartile is farther from the median than the upper quartile.
But the problem is when I use another method to determine skewness:
mean (337.5) > median (325)
This indicates the data is positively skewed.
Did I miss something?
7,93552348
One measure of skewness is based on mean-median - .
Another measure of skewness is based on the relative quartile differences (Q3-Q2) vs (Q2-Q1) expressed as a ratio
When (Q3-Q2) vs (Q2-Q1) is instead expressed as a difference (or equivalently midhinge-median), that must be scaled to make it dimensionless (as usually needed for a skewness measure), say by the IQR, as
(by putting $u=0.25$).
The most common measure is of course .
There's no reason that these three measures will necessarily be consistent. Any one of them could be different from the other two.
What we regard as "skewness" is a somewhat slippery and ill-defined concept. See
for more discussion.
If we look at your data with a normal qqplot:
[The line marked there is based on the first 6 points only, because I want to discuss the deviation of the last two from the pattern there.]
We see that the smallest 6 points lie almost perfectly on the line.
Then the 7th point is below the line (closer to the middle relatively than the corresponding second point in from the left end), while the eighth point sits way above.
The 7th point suggests mild left skew, the last, stronger right skew. If you ignore either point, the impression of skewness is entirely determined by the other.
If I had to say it was one or the other, I'd call that "right skew" but I'd also point out that the impression was entirely due to the effect of that one very large point. Without it there's really nothing to say it's right skew. (On the other hand, without the 7th point instead, it's clearly not left skew.)
We must be very careful when our impression is entirely determined by single points, and can be flipped around by removing one point. That's not much of a basis to go on!
I start with the premise that what makes an outlier 'outlying' is the model (what's an outlier with respect on one model may be quite typical under another model).
I think an observation at the 0.01 upper percentile (1/10000) of a normal (3.72 sds above the mean) is equally an outlier to the normal model as an observation at the 0.01 upper percentile of an exponential distribution is to the exponential model. (If we transform a distribution by its own probability integral transform, each will go to the same uniform)
To see the problem with applying the boxplot rule to even a moderately right skew distribution, simulate large samples from an exponential distribution.
E.g. if we simulate samples of size 100 from a normal, we average less than 1 outlier per sample. If we do it with an exponential, we average around 5. But there's no real basis on which to say that a higher proportion of exponential values are "outlying" unless we do it by comparison with (say) a normal model. In particular situations we might have specific reasons to have an outlier rule of some particular form, but there's no general rule, which leaves us with general principles like the one I started with on this subsection - to treat each model/distribution on its own lights (if a value isn't unusual with respect to a model, why call it an outlier in that situation?)
To turn to the question in the title:
While it's a pretty crude instrument (which is why I looked at the QQ-plot) there are several indications of skewness in a boxplot - if there's at least one point marked as an outlier, there's potentially (at least) three:
In this sample (n=100), the outer points (green) mark the extremes, and with the median suggest left skewness. Then the fences (blue) suggest (when combined with the median) suggest right skewness. Then the hinges (quartiles, brown), suggest left skewness when combined with the median.
As we see, they needn't be consistent. Which you would focus on depends on the situation you're in (and possibly your preferences).
However, a warning on just how crude the boxplot is. The example toward the end
-- which includes a description of how to generate the data --
gives four quite different distributions with the same boxplot:
As you can see there's a quite skewed distribution with all of the above-mentioned indicators of skewness showing perfect symmetry.
Let's take this from the point of view "what answer was your teacher expecting, given that this is a boxplot, which marks one point as an outlier?".
We're left with first answering "do they expect you to assess skewness including that point, or with it in the sample?". Some would exclude it, and assess skewness from what remains, as jsk did in another answer. While I have disputed aspects of that approach, I can't say it's wrong -- that depends on the situation. Some would include it (not least because excluding 12.5% of your sample because of a rule derived from normality seems a big step*).
* Imagine a population distribution which is symmetric except for the far right tail (I constructed one such in answering this - normal but with the extreme right tail being Pareto - but didn't present it in my answer). If I draw samples of size 8, often 7 of the observations come from the normal-looking part and one comes from the upper tail. If we exclude the points marked as boxplot-outliers in that case, we're excluding the point that's telling us that it is actually skew! When we do, the truncated distribution that remains in that situation is left-skew, and our conclusion would be the opposite of the correct one.
102k12150329
No, you did not miss anything: you are actually seeing beyond the simplistic summaries that were presented.
These data are both positively and negatively skewed (in the sense of "skewness" suggesting some form of asymmetry in the data distribution).
John Tukey described a systematic way to explore asymmetry in batches of data by means of his "N-number summary."
A boxplot is a graphic of a 5-number summary and thereby is amenable to this analysis.
A boxplot displays a 5-number summary: the median $M$, the two hinges $H^{+}$ and $H^{-}$, and the extremes $X^{+}$ and $X^{-}$.
The key idea in Tukey's generalized approach is to choose some statistics $T_i^{+}$ reflecting the upper half of the batch (based on ranks or, equivalently, percentiles), with increasing $i$ corresponding to more extreme data.
Each statistic $T_i^{+}$ has a counterparts $T_i^{-}$ obtained by computing the same statistic after turning the data upside-down (by negating the values, for instance).
In a symmetric batch, each pair of matching statistics must be centered at the middle of the batch (and this center will coincide with $M = M^{+}=M^{-}$).
Thus, a plot of how much the mid-statistic $(T_i^{+} + T_i^{-})/2$ varies with $i$ provides a graphical diagnostic and can furnish a quantitative estimate of asymmetry.
To apply this idea to a boxplot, just draw the midpoints of each pair of corresponding parts: the median (which is already there), the midpoint of the hinges (ends of the box, shown in blue), and the midpoint of the extremes (shown in red).
In this example the lower value of the mid-hinge compared to the median indicates the middle of the batch is slightly negatively skewed (thereby corroborating the assessment quoted in the question, while at the same time suitably limiting its scope to the middle of the batch) while the (much) higher value of the mid-extreme indicates the tails of the batch (or at least its extremes) are positively skewed (albeit, on closer inspection, this is due to a single high outlier).
Although this is almost a trivial example, the relative richness of this interpretation compared to a single "skewness" statistic already reveals the descriptive power of this approach.
With a small amount of practice you do not have to draw these mid-statistics: you can imagine where they are and read the resulting skewness information directly off any boxplot.
An example from Tukey's EDA (p. 81) uses a nine-number summary of heights of 219 volcanoes (expressed in hundreds of feet).
He calls these statistics $M$, $H$, $E$, $D$, and $X$: they correspond (roughly) to the middle, the upper and lower quartiles, the eighths, the sixteenths and the extremes, respectively.
I have indexed them in this order by $i=1, 2, 3, 4, 5$.
The left hand plot in the next figure is the diagnostic plot for the midpoints of these paired statistics.
From the accelerating slope, it is clear the data are becoming more and more positively skewed as we reach out into their tails.
The middle and right plots show the same thing for the square roots (of the data, not of the mid-number statistics!) and the (base-10) logarithms.
The relative stability of the values of the roots (notice the relative small vertical range and the level sloped in the middle) indicates that this batch of 219 values becomes approximately symmetric both in its middle portions and in all parts of its tails, almost out to the extremes when the heights are re-expressed as square roots.
This result is a strong--almost compelling--basis for continuing further analysis of these heights in terms of their square roots.
Among other things, these plots reveal something quantitative about the asymmetry of the data: on the original scale, they immediately reveal the varying skewness of the data (casting considerable doubt on the utility of using a single statistic to characterize its skewness), whereas on the square root scale, the data are close to symmetric about their middle--and therefore can succinctly be summarized with a five-number summary, or equivalently a boxplot.
The skewness again varies appreciably on a log scale, showing the logarithm is too "strong" a way to re-express these data.
The generalization of a boxplot to seven-, nine-, and more-number summaries is straightforward to draw.
Tukey calls them "schematic plots."
Today many plots serve a similar purpose, including standbys like Q-Q plots and relative novelties such as "bean plots" and "violin plots."
(Even the lowly histogram can be pressed into service for this purpose.)
Using points from such plots, one can assess asymmetry in a detailed fashion and perform a similar evaluation of ways to re-express the data.
108k9191384
The mean being less than or greater than the median is a shortcut that often works for determining the direction of skew so long as there are no outliers. In this case, the distribution is negatively skewed but the mean is larger than the median due to the outlier.
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