# Thread: normal distribution help

1. ## normal distribution help

Jam is sold in jars and the mean weight of the contents is 108 grams. Only 3% of jars have contents weighing less than 100 grams. Assuming that the weight of jam in a jar is normally distributed find
a the standard deviation of the weight of jam in a jar,
b the proportion of jars where the contents weigh more than 115 grams.
J∼N(108,σ2)

a) $\displaystyle p(J<100) = 0.03$

From here do I not just look up the z value for 0.03 in the 'z tables'? In the book it gives z = -1.88? I dont know how they got this number. Can anyone explain?

I thought the equation should be $\displaystyle Z = \frac{100-108}{\sigma} = 0.5120$

0.5120 is the value in the 'z'' tables that corresponds to 0.03.

but the correct equation to solve for the standard deviation is $\displaystyle \frac{100-108}{\sigma} = -1.88$

Thanks.

2. you should transform J into standard normal distribution variable before solving the equation for $\displaystyle {\sigma}$

which is $\displaystyle \frac{J-108}{\sigma}$