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Given that 40% of our toys weigh between $\mu$ and $\mu + 30$ where $\mu = 230 \text{grams}$ and their weight is distributed normally. Then what is the Standard deviation ?

My way:

I moved to standard normal dist. $ Z = \frac{X - 230}{ s}$ and know that $$ \Phi(\frac{260}{s}) - \Phi(0)= 0.4$$

So:

$$ \Phi( \frac{260}{s}) = 0.9 \\ \frac{260}{s} \approx 1.28 \\ s \approx 203.125$$

But I am not sure at all, and furthermore, how do I find the exact number and not the approximation? because I looked in the table and only found the closest to 0.9 .... is there a way? (I need to calculate like this) Thank you

CSch of x
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  • I think this is the way to go... you just need to review your calculations. If you were able to compute it exactly, the same procedure would allow you to compute exactly any value of the normal distribution. – PierreCarre Oct 12 '20 at 14:52

1 Answers1

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be careful....

$\Phi^{-1}(0.89973)=1.28$ is a good approximation of $\Phi^{-1}(0.9)$ .

The problem of your solution is another one. You found that $\sigma \approx 203$. This means that $\mu \pm 3\sigma$ is an interval like this $[-379;839]$ with a lot of toys with negative weight...

Of course the correct solution is this:

$$\Phi\Bigg(\frac{260-230}{\sigma}\Bigg)-\Phi\Bigg(\frac{230-230}{\sigma}\Bigg)=0.4$$

$$\Phi\Bigg(\frac{30}{\sigma}\Bigg)-\Phi(0)=0.4$$

$$\Phi(\frac{30}{\sigma})-0.5=0.4$$

$$\Phi\Bigg(\frac{30}{\sigma}\Bigg)=0.9$$

$$\frac{30}{\sigma}\approx 1.28$$

$$\sigma \approx 23.41$$

tommik
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