I just need to know how to get my z score
the answer is z = 1.281, z = -1.281.

but how did they get that ???

orignal quistion
a commercial prawn fisherman recorded the number of prawns he took from each trap. The average number of prawns per trap was 230 with a standard deviation of 22. What number of prawns per trap would you expect in the interval symmetrical about the mean where 80% of the numbers would be found?

thanks for the help :)

Cola

Question

I just need to know how to get my z score 
the answer is  z = 1.281, z = -1.281.

but how did they get that ???

orignal quistion 
 a commercial prawn fisherman recorded the number of prawns he took from each trap. The average number of prawns per trap was 230 with a standard deviation of 22. What number of prawns per trap would you expect in the interval symmetrical about the mean where 80% of the numbers would be found?

thanks for the help :)

Cola

jim_thompson5910 · Answer

You have a standard normal distribution where you want to find the z-score where 80% of the distribution lies between z = -k and z = k for some positive value of k

So if you draw it out, you'll get something like this


|dw:1339618718646:dw|