Find the standard deviation of the weights of the bags of popcorn (S1 Edexcel Help)

The weights of bags of popcorn are normally distributed with mean of 200 g and 60% of
all bags weighing between 190 g and 210 g.

I've read the mark scheme and I still don't get it. Could someone go through it step by step for me please?
If it's helps its from the January 2008 Edexcel S1 paper, found here:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202008/6683_01_que_20080115.pdf

Ok, the first thing you need to do is find z. What we know about a normal distribution is that it is symmetrical. Therefore 30% of the data lies between 210 and 200. Also we know that 50% of the data lies below the mean. So 80% of the data lies below 210. So you look on your statistical tables for a 0.8 level of probability which come out as z=0.842.

So z=0.842
The equation used to derive z is (value[i.e.210] - mean)/standard deviation = z
Therefore 210-200/s = 0.842
Therefore 10=0.842s
Therefore s=10/0.842 = 11.876

I don't actually know what the answer is so I don't know if it's right but I hope it helped anyway.

You need a table of the standard normal distribution to find out how many SDs are in the range that encompasses 60% of the sample.

As I am sure you know, one SD either side of the mean encompasses ~68.2% of the sample (~34.1% on either side), so it is obvious that the SD of the weights is a bit more than 10.

See the table at http://www.mathsisfun.com/data/standard-…

a one-side difference from the mean of 30% is found at SD = 0.84 = 10 and therefore SD = 11.9

Understanding and using the various forms of the standard normal distribution table is probably one of the most important things to understand about elementary statistics

(mean) μ = 200 g
(z-score) z = 0.84 (found by looking for the closest to 0.3000 on the chart)
x = 190, 210

z = |μ - x| / σ
→ σ = |μ - x| / z

σ = |200 - 190| / 0.84
σ ≈ 11.9