The weight in grams, of beans in a tin is normally distributed with mean U and S.D. 7.8, given that 10% conntain more than 225g a) Find U b) % of tins that contain more than 225 grams(A2 stats)

for part a) you would need to use the formula (x-U)/S.D.=a where X=200 and S.D.=7.8. The student would need to calculate a with the information given. I here would use standard normal distribution here to find a value for a, Z-N(0,1^2). Typing this into a calculator with an area of 0.1 gives the value of -1.2816. I would emphasise the negative sign which can be inferred from a diagram I would draw whilst demonstarting the question. Then, with this infomration I would use the equation I quoted at the beginning and plug all the numbers in as X=200, U=?, S.D.=7.8 and a=-1.2816. Here there is one unknown and you would re-arrange and solve.
For part b) I would use the information I worked out in part a), as with the nature of our new spec, this whole question can be done entirely by our calculators, and with time pressures in exam I would emphasise to the student there is no need to spend too long on this question at all. You would just plug in an lower bound (223) and an upper bound (any large number) and input the S.D. and U and multiple the value calculated for the percentage answer.