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OK. Sorry if this is a stupid question, I'm just a little rusty on my statistics.

I have the mean and standard deviation of a data set. The data is a normal with possibilities $\{1,2,3,4,5\}$ with $n \approx 40$.

I am looking to find what percentage of people answered 4 or 5.

How would I compute that number?

Thanks!

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Do you know anything else? In particular, the distribution of the results? You can't actually answer that question without it... –  David Z Dec 12 '11 at 0:28
    
@DavidZaslavsky Normal –  Nick Dec 12 '11 at 0:29
    
@mixedmath it's normal. I'm sorry for not knowing correct terminology –  Nick Dec 12 '11 at 0:36
    
@Nick: That's fine, but perhaps you can tell me more. What does it mean for a (discrete) set to have a normal distribution? How many people answered 3, for instance? –  mixedmath Dec 12 '11 at 0:40
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@nick what mixedmath is saying is that saying something is Normal implies that the data is continuous. Saying a discrete sample space is Normal has little meaning. Moreover, even if your data were continuous, saying they're normal without specifying the parameters (mean and SD) has even less meaning. Perhaps you could supply a bit more context? Where did this problem come from? How do you know the data are well approximated by a normal distribution? What are the parameters of that distribution? –  Drew Christianson Dec 12 '11 at 0:44
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5 Answers

As I mentioned in my comment, you can't really answer this question without knowing the distribution of the results, not just the mean and standard deviation. Let's say your values are distributed according to the binomial distribution (which is the discrete analog of the normal distribution) for a 4-trial binary experiment. The probability of the result being $k + 1$ is

$$P(k + 1) = \binom{4}{k}p^k(1-p)^{4-k}$$

But you don't have the value of $p$ directly; instead, you have the mean $\mu + 1 = 4p$ and standard deviation $\sigma + 1 = \sqrt{4p(1-p)}$. (The +1's come up because your data start at 1 rather than 0.) So you will need to solve for $p$ as $p = \frac{\mu + 1}{4}$. You should check that plugging this into the standard deviation formula $\sqrt{4p(1-p)}$ gives you something close to the actual standard deviation of your data, otherwise the binomial distribution is not an accurate representation of your data.

Once you have satisfied yourself that the binomial distribution is roughly accurate and that you have properly calculated $p$, you can compute the probability of getting a 4 or a 5 as

$$P(4) + P(5) = 4p^3(1-p) + p^4$$

This will give you a good approximation to the fraction of 4s and 5s in your data set; just multiply by $n$ to get the number of 4s and 5s. How good the approximation is depends on how well your data fit the binomial distribution, and you can use the matching of the standard deviations as an indicator for that.

I don't know of any practically useful way to get the number of 4s and 5s more exactly. You could write out the equations for the mean and standard deviation,

$$\begin{align}\frac{1}{n}\sum_{i=1}^n k_i &= \mu & \frac{1}{n}\sum_{i=1}^n (k_i - \mu)^2 &= \sigma^2\end{align}$$

and try to plug in numbers for the $k_i$ to see if you can come up with a set that matches your $\mu$ and $\sigma$ exactly, but that's just trial and error and it will likely not have a unique solution anyway.

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If you told us the mean and standard deviation, we could tell you the limits on the proportion having the values $4$ or $5$.

Here is an example: suppose the mean was $2.5$ and the standard deviation was $1$. There are various possible probability distributions with this result and this table shows two of them.

Answer Prob1  Prob2 
  1     9/32   1/12
  2      0     5/8 
  3    11/16    0  
  4      0     7/24    
  5     1/32    0   

In the first case the proportion choosing 4 or 5 is $0.03125$; in the second it is $0.291666\ldots$. Any figure between these is also possible for this particular mean and standard deviation. If the distribution had to be unimodal then the limits would be tighter.

The upper and lower limits on the proportion choosing 4 or 5 would increase if the mean or the standard deviation were higher.

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Suppose you have $n = n_1 + n_2 + n_3 + n_4 + n_5$ data points where $n_i$ data points have value $i$ for $i \in \{1,2,3,4,5\}$. The probabilistic version of this is a discrete random variable taking on values $\{1,2,3,4,5\}$ with probabilities $p_i = n_i/n$, and of course $p_1 + p_2 + p_3 + p_4 + p_5 = 1$. You know the mean $\mu$ and standard deviation $\sigma$ of this random variable. Obviously. $1 \leq \mu \leq 5$. Less well known is the result that $0 \leq \sigma \leq 2$. The extreme value $\sigma = 0$ occurs when one of the $p_i$'s equals $1$ and all other $p_j$ are zero, that is, all the data points have value $i$ (and so $\mu = i$), while the other extreme value $\sigma = 2$ occurs when $p_1 = p_5 = \frac{1}{2}, p_2 = p_3 = p_4 = 0$. Note that $\mu = 3$ in this case.

Turning to your problem of estimating $p_5$, we have that $$1^2\cdot p_1 + 2^2\cdot p_2 + 3^2\cdot p_3 + 4^2\cdot p_4 + 5^2\cdot p_5 = \sigma^2 + \mu^2$$ where the right side is known to you. Since $$1^2\cdot p_1 + 2^2\cdot p_2 + 3^2\cdot p_3 + 4^2\cdot p_4 > p1 + p_2 + p_3 + p_4 = 1-p_5$$ we have a crude upper bound $$p_5 \leq \frac{\sigma^2 + \mu^2 - 1}{24}.$$ Note that in the extreme case $\sigma = 0$, the bound is $(\mu^2 - 1)/24$ which is exact if all the data points have value $5$ (or $1$) but not very good when all the data points have some other value and $p_5 = 0$. But, if you know that $\sigma = 0$, you shouldn't be using the bound anyway because you know that $p_\mu = 1$ and all other $p_j$ are $0$. In the other extreme case $\sigma = 2, \mu = 3$, the upper bound is $\frac{12}{24} = \frac{1}{2}$ and is exact. But again, if you know that $\sigma = 2$, you shouldn't bother with the bound since this extreme distribution is known exactly.

Another weak bound comes from the one-sided Chebyshev inequality which gives $$p_5 \leq \frac{\sigma^2}{\sigma^2 + (5-\mu)^2}.$$ Which of the two bounds described here is tighter depends on the values of $\mu$ and $\sigma$.

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Additionally, there is the option of using the Central Limit Theorem.

What you're really looking for is the probability that a given reply $X_i$ is 4 or more:$P(X_i\geq 4)$. Multiply that by the sample size, and you have a reasonable estimate of the number of responses that were 4 or 5. So, we use the CLT to approximate the data.


EDIT: Per Henry and Dilip's comments it you should us a continuity correction at this point. So, we rewrite $P(X\geq 4)$ as $P(X >3.5)$. This corrects for some of the error inherent in using the normal distribution to approximate a binomial.


What the CLT says is that, if $X_i$ obey any distribution with mean $\mu$ and variance $\sigma^2$ (note that $SD(X) = \sqrt{Var(X)}$), then $\frac{\overline{X_n}-\mu}{\frac{\sigma}{\sqrt{n}}}$ where $\overline{X_n}$ is the mean of a sample of size n is well approximated by a normal distribution with mean 0 and variance 1. So, we set about transforming the probability above into the form necessary for the CLT:$$ P(X_i> 3.5) = P\left(\sum_{i=1}^{40}X_i > \sum_{i=1}^{40}3.5\right) = P\left(\frac{\sum_{i=1}^{40}X_i}{40}> \frac{\sum_{i=1}^{40}3.5}{40}\right) = P(\overline{X_{40}}>3.5) = P(\overline{X_{40}}-\mu> 3.5-\mu) $$$$P\left(\frac{\overline{X_{40}}-\mu}{\frac{\sigma}{\sqrt{40}}}> \frac{3.5-\mu}{\frac{\sigma}{\sqrt{40}}}\right) = P\left(Z> \frac{3.5-\mu}{\frac{\sigma}{\sqrt{40}}}\right) \;\;\;\;\;\;\;Z\thicksim Normal(0,1)$$ Plug in values of $\sigma$ and $\mu$ appropriate to the problem, and you can approximate the probability using a normal distribution.

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If there are only five possible answers then you might want a continuity correction for your Normal approximation, e.g. by looking at $\Pr(X_i \ge 3.5)$ –  Henry Dec 12 '11 at 1:56
    
@henry I'm not familiar with that adjustment, or at least the term. I'm only just finishing up a probability theory course. Link? –  Drew Christianson Dec 12 '11 at 2:10
    
    
@DrewChristianson "I'm only just finishing up a probability theory course. " Since you claimed here that you are using Ross's book in your course, look at page 205 in the 8th edition. –  Dilip Sarwate Dec 12 '11 at 17:53
    
@dilip thanks for the reference, I appreciate it. –  Drew Christianson Dec 12 '11 at 18:15
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My question is if I have A mean of $ 30,000 an of $3,00 satandard diviation. It is base with the annual incomes of some students. I need to find the percent of students who earned more than $36,000.

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This is not an answer to the question. –  Asaf Karagila Feb 19 '13 at 23:50
    
This is not an answer to the question. If you have a question, please use the "Ask Question" link near the top right of the page. –  robjohn Feb 20 '13 at 0:08
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