How to estimate the range of a normal distribution when the mean and standard deviation are given

Question

For example, how would you respond to this question?

The earnings of one-hundred workers in a company are normally distributed. If the mean of this data set is 24 and the standard deviation is 4, find an approximate value for the range.

Answer 1

Recall that about $99.7\%$ of data under a normal curve falls within three standard deviations of the mean. When you are given the mean and standard deviation, this seems like a pretty good way to approximate the range. So since the mean is $24$, we could estimate that most of the data falls in the interval $$[24-3(4), \;24+3(4)] \;=\; [12,36].$$ So the range is $36-12 = 24$.

Answer 2

For an independent sample of size $N$ from a continuous random variable $X$ with CDF, the CDF's for the maximum and minimum of the sample are $$ \eqalign{\mathbb P(X_{max} \le r) &= F(r)^N\cr \mathbb P(X_{min} \le r) &= 1 - (1-F(r))^N\cr}$$ In particular, if $N=100$, with probability $0.9$ $X_{max}$ is in the interval where $F(r)$ goes from $0.05^{1/100} \approx 0.970487$ to $0.95^{1/100} \approx 0.999487$, and with probability $0.9$ $X_{min}$ is in the interval where $F(r)$ goes from $1-0.95^{1/100} \approx 0.000513$ to $1-0.05^{1/100} \approx 0.029513$. $X_{max}$ and $X_{min}$ are not independent, but we can say that with probability at least $0.8$ both of these are true and the range is between $F^{-1}(0.970487) - F^{-1}(0.029513)$ and $F^{-1}(0.999487) - F^{-1}(0.000513)$. In the case of a normal distribution with mean $\mu$ and standard deviation $\sigma$, that corresponds to a range of between approximately $3.776 \sigma$ and $6.567 \sigma$.

I don't know of a closed-form result for the distribution of the range in the normal case, but in a simulation of $10000$ such samples, the median range was $4.9514 \sigma$, with $10$'th percentile $4.2748 \sigma$ and $90$'th percentile $5.8001 \sigma$.

So we can't really give a "good approximation" for the range, but a reasonable guess is about $5 \sigma$.

Of course the statement that the earnings are normally distributed is not literally true (unless negative earnings are possible and irrational earnings are almost certain). What might be true is that the earnings distribution is well approximated in some sense by a normal distribution. Usually such approximations may be pretty good near the middle of the distribution, but bad in the tails (nobody has negative earnings, and the CEO may be many standard deviations above the mean). Unfortunately, the range is very sensitive to the tails, so the normal approximation may not be very good in practice.

Answer 3

The answer is 24. Recall normal distribution graph.

• Around 68% of data is between standard deviation -1 to 1. Here it is between 20(24-1x4) to 28(24+1x4).
• Around 96% of data is between standard deviation -2 to 2. Here it is between 16(24-2x4) to 32(24+2x4).
• Around 100% of data is between standard deviation -3 to 3. Here it is between 12(24-3x4) to 36(24+3x4).

Range = Maximum number - Minimum number = 36-12 = 24

Answer 4

This is an old question (asked about 5 years ago), but none of the answers given sound right to me. Any answer should explain that the range is larger for larger sample sizes. (More data points will result in a wider range.)

The answer is found using a Normal Probability Plot. (See wikipedia.org for a description.)

n is the number of sample points. For n>10, the probability of the lowest value is 0.5/n. Calculate the normal inverse using probability 0.5/n, mean 24, standard deviation 4. In Excel this is "low = NORMINV(0.5/n,24,4)" The problem statement is symmetric, so the high is reflected about the mean: "high = 48 - low". Range = high - low.

The expected range of 100 samples (mean 24, standard deviation 4) is about 21. The actual range of any one experiment will be different, depending on chance. The expected range increases to about 39 with 1 million samples.

    n      low   high  range
    10     17.4  30.6  13.2
   100     13.7  34.3  20.6
  1,000    10.8  37.2  26.3
  10,000   8.4   39.6  31.1
 100,000   6.3   41.7  35.3
1,000,000  4.4   43.6  39.1

The problem statement says "the earnings ... are normally distributed". What does it mean to be "normally distributed"? The normal distribution is a probability distribution. So one requirement of normality is the lowest earning must match a normal curve with probability 0.5%, mean 24, standard deviation 4. A Normal Probability Plot of the earnings must be a straight line.