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using a z score
\[z =(x - \mu)/\sigma\]
\[z = (160 - 150)/35\]
z = 0.2857

@campbell_st actually it should be below 50% |dw:1335172829891:dw|

I don't know which is the right way to look at this problem.

DBhatta's graph sums it up

oh, is what I have done a z-test instead of a normal distribution?

yes my friend. It should be normal distribution \[z = x - u / \sigma \]

But it has that 40 accounts info, isn't that n=40?

ah...let me see. it's becoming more interesting now.

haha...okay, thanks!

oh yes, it's z-distribution.

oh so we should use the other method? The one I written just now? But why?

Yes use the one you have written. We are assuming that population mean is equal to the sample mean.

oh I see...Thanks for you help!! :)

Actually you helped me, I was off-track. It was you who made me realize. :)