Look up the value of z that is exceeded 20% of the time for a standard

normal distribution. This is ~0.84.

Now if X~N(mu,sigma^2), then z=(x-mu)/sigma has a standard normal distribution. So (x-mu)/sigma > 0.84 20% of the time, rearrange:

x>mu+(0.84*sigma)

20% of the time. But we are told that the value exceeded 20% of the time is

23.3, so:

23.3=mu+0.84*sigma,

and mu=22.8, so:

23.3=22.8+0.84*sigma,

so:

sigma=0.5/0.84=0.5952..

Which is the standard deviation, so the variance is sigma^2~=0.3543

With these figures the z corresponding to 21.82 is:(b) The record over this distance for women at her club is 21.82 seconds. According to her

model, what is the chance that she will beat this record in her next race?

(3 marks)

z=(21.82-22.8)/0.5952 ~= -1.647

Looking this up in the normal table the probability of logging a time less than 21.82

is ~=0.0495.

RonL