Let the weight of the chickens be . Then
where
is the standard deviation of the weights and
is the mean of the weights. Then, the statement you're given is
.
Now, rewrite where
. Then, look up
in a standard normal table such that
and then note
. Solve for
, and you are done.
Statistics - Calculating Mean given standard deviation and percentage. - Mathematics Stack Exchange
t test - How to calculate standard deviation when only mean of the data and sample size is available? - Cross Validated
What's the best way to calculate standard deviation in this problem? From October 20222
ELI5: Standard deviation
How is standard deviation calculated?
Why do we calculate standard deviation?
What is the formula to calculate standard deviation?
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Let the weight of the chickens be . Then
where
is the standard deviation of the weights and
is the mean of the weights. Then, the statement you're given is
.
Now, rewrite where
. Then, look up
in a standard normal table such that
and then note
. Solve for
, and you are done.
You have to use the fact that the weights are normally distributed.
You're given the standard deviation, and a value for the left 4% of the distribution.
You can calculate (or look up) how many standard deviations away from the mean you are at 4%.
Given this, you multiply the number of standard deviations from the mean by the standard deviation (in lbs) to see how many lbs away from the mean you are.
You know one end (20 lbs) and you can now find the other end, which is the mean.
It's impossible. Consider any vector that has a mean of $0$ - multiply the values by $100$, and the standard deviation also increases by a factor of $100$, but the mean and sample size are unchanged. Both vectors have the same mean and $n$, but different SD. In general, you cannot find the standard deviation of the data given only the mean value and number of data points.
In general this is not possible, as noted in the other answers.
But, if you know (or assume) something about the underlying distribution from which the data points have been drawn, then it is sometimes possible.
The following table shows distributions where it is indeed relatively easy to do so. Shown are the parameters, mean $\mu$ and variance as given in the different Wikipedia links, as well as the standard deviation $\sigma = \sqrt{Var}$ with the mean substituted in and simplified.
| Distribution | Parameters | Mean $\mu$ | Variance | Standard Deviation |
|---|---|---|---|---|
| Poisson | $\lambda \ge 0$ | $\lambda$ | $\lambda$ | $\sqrt{\mu}$ |
| Exponential | $\lambda > 0$ | $1/\lambda$ | $1/\lambda^2$ | $\mu$ |
| Binomial | $n \in \mathbb{N}, 0 < p < 1$ | $np$ | $np(1-p)$ | $\sqrt{\mu - \mu^2/n}$ |
| (Central) chi-squared | $k \in \mathbb{N}$ | $k$ | $2k$ | $\sqrt{2\mu}$ |
| Maxwell-Boltzmann | $a>0$ | $2a\sqrt{2/\pi}$ | $\frac{a^2(3\pi-8)}{\pi}$ | $\mu\sqrt{\frac{3}{8}\pi-1}$ |
You may note that most of these distributions depend on only one parameter. I would conjecture that actually most single-parameter distributions for which closed-form solutions for both mean and variance exist allow you to deduce the standard deviation from the mean. So this table is probably incomplete.
Conversely, if the distribution depends on two parameters, it is quite unlikely that there is a simple relationship between mean and variance.
The following table shows some distributions where it is not possible to deduce the standard deviation just from the mean (and the sample size). Since many distributions fall into that category, this table is definitely not complete.
| Distribution | Parameters | Mean $\mu$ | Variance |
|---|---|---|---|
| Uniform | $a < b$ | $\frac{1}{2}(a+b)$ | $\frac{1}{12}(b-a)^2$ |
| Normal | $\mu, \sigma$ | $\mu$ | $\sigma^2$ |
| Student's-t | $\nu > 0$ | 0, for $\nu > 1$ | $\nu/(\nu-2)$, for $\nu > 2$ |
| Beta | $\alpha > 0, \beta > 0$ | $\alpha/(\alpha + \beta)$ | $\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$ |
| Gamma | $\alpha>0, \beta>0$ | $\alpha/\beta$ | $\alpha/\beta^2$ |
Note in particular the Student's t-distribution, which may be relevant due to the t-test tag on the question.