# Could someone explain a AS Statistics exam question to me?

Its question 7c and d on the January 2005 paper, I found the mark scheme by googling it but it is very vague on its calculation method. This would be a tremendous help, thank you

(Edexcel Statistics)

Posted Fri 12th April, 2013 @ 20:29 by Cymbeline

Part (c):

You are told that P(X 79 + b) = 2P(X 79 – a)

Which means the small area to the right is twice the size of the small area to the left of the graph.

The area of the shaded region is basically the probability P(X 79 - a)

The area of the whole normal distribution graph is 1.

To get the joint area of the two smaller regions of the graph (the left and right):

Joint area = 1 – large region of the graph

Joint area = 1 - 0.6463 = 0.3537

The joint area is three times the size of the shaded area on the left

So the shaded area = 1/3 of the joint area = 1/3 (0.3537) = 0.1179

Part (d):

The area less than 79 + b: P(X 79 + b) = 0.1179 + 0.6463 = 0.7642

You should know the equation z = (x – μ) / σ

z = (79 + b) – 79 / 12 (σ = root 144)

z = b/12

b/12 = 0.72 (from table in formula book)

b = 8.64

Answered Sat 13th April, 2013 @ 22:54 by Charlotte