+118608 Have you been given a table?
Maybe like this one
Yours may be different presented differenly.
If 5.48% contain more, then (100-5.48)% = 94.52% = 0.9452
Find 0.9452 in the body of the table.
I highlighted it, it is 1.60
So 14.5oz has a Z-score of 1.6
the average is 14.1 oz with a z-score of 0
0.4 oz => z-score of 1.6
0.4/1.6=0.25
So the mean of your distribution is 14.1 oz and the standard deviation is 0.25g
B)
14.1-13.9=0.2 grams below the mean. 0.2
13.9 grams is 0.2 grams below the mean
0.2/0.25=0.8
That is 0.8 of a standard deviation.
So the z score will be -0.8
the area below -0.8 is the same as the area above +0.8.
Which is also the same as 1- (the area below +0.8)
Which from the table is 1-0.7881 = 0.2119.
So 21.19% will be rejected.
Here is a simple little site where you can check your answers
http://davidmlane.com/hyperstat/z_table.html
C very high.
Yes I know you are meant to do more than that. But that is enough for me.
Out of 200 I'd expect about 40 to be discarded.
To me this seems to be a very unrealistic value but I think that is what this question predicts.
how do I do this?
