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"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.

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1 Answer 1

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Hint:

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

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