Finding SD by only knowing the mean and one other piece of information

Question

"The weights of bags of vegetables are approximately normally distributed with mean of 1kg. One quarter have weight greater than 1.1kg. Find the standard deviation."

I know this question might be simple to most of you, but probability/statistics really isn't my strong suit. I usually understand the standard deviation to be $\sigma=\sqrt{\sum(x_i-\mu)^2\cdot N^{-1}}$. In this case we aren't concerned with specific values of $x$ I don't think, and we don't know the sample size. All I can extrapolate from the question is that since it's 25% that have weight greater than 1.1kg then that 25% starts less than one standard deviation away from the mean.

If I were to guess, I would say that since this is normally distributed then I know that 34% of the bags fall within one standard deviation above the mean, and if 25% have a weight greater than 1.1kg, then 25% have a weight between 1kg and 1.1kg. So the deviation for 25% is 0.1kg which is less than 1 standard deviation, so I think I should multiply 0.1 by $\frac{34}{25}$ to get $0.136$ which, if my assumptions are correct, is equal to one standard deviation?

This doesn't feel concrete, I'm sure there's a more rigorous way to find the answer, any help is appreciated.

Answer 1

All I can extrapolate from the question is that since it's 25% that have weight greater than 1.1kg

That´s true. So the equation $P(X\geq 1.1)=0.25$, where $X\sim \mathcal N(1, \sigma^2)$ Next we can standardize the random variable: $Z=\frac{X-\mu}{\sigma}$

So the equation is $P\left(Z\geq \frac{X-\mu}{\sigma}= \frac{1.1-1}{\sigma}\right)=0.25$. Thus

$$1-\Phi\left(\frac{0.1}{\sigma} \right)=0.25$$ Can you proceed?

Answer 2

One quarter have weight greater than 1.1kg

I translate it for you: $75\%$ of bags has a weight that is less or equal to $1.1$ kg. Assuming normality with mean 1 this can be formalized in the following way

$$\Phi\Bigg(\frac{1.1-1}{\sigma}\Bigg)=0.75$$

$$\frac{1.1-1}{\sigma}\approx 0.6745$$

$$\sigma\approx 0.1483$$

Answer 3

Given that the normal distribution is symmetric and that 1/4 of the vegetables have weight larger than 1.1 you can say:

$\int_{-\infty}^{0.9} f(x)· \,dx = 1/4 $

$\int_{0.9}^{1} f(x)· \,dx = 1/4 $

$\int_{1}^{1.1} f(x)· \,dx = 1/4 $

$\int_{1.1}^{+\infty} f(x)· \,dx = 1/4 $

The inter-quantile range (IQR) is defined as:

$IQR = Q_{75} - Q_{25} = 1.1 - 0.9 = 0.2$

It can be shown that for a normal distribution:

$IQR = 2 ·\phi^{-1}(0.75)·\sigma$

Where $\phi^{-1}$ is the inverse cumulative density function or quantile function of the standard normal distribution. Hence:

$\sigma = \frac{IQR}{2 ·\phi^{-1}(0.75)} \approx \frac{0.2}{2 · 0.6745} = 0.1483 $

To check the answer, you can ran in R:

x = rnorm(10000000, 1, 0.2/(2*qnorm(0.75)))
plot(density(x))
quantile(x)
# 0%       25%       50%       75%      100% 
# 0.2135559 0.9000991 1.0001097 1.0999769 1.7831438