"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.


1 Answer 1



That is not quite the intended approach.

Instead you should say from the standard normal distribution that the upper quartile point is $\Phi^{-1}(0.75) \approx 0.6745$, i.e. $25\%$ of a normal distribution is more than about $0.6745$ standard deviations above the mean.

You then calculate the standard deviation by saying here $0.6745$ standard deviations is $1.1-1.0=0.1$kg and doing the division


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.