Suppose that in a particular year the percentage increase in the value of stocks listed on the Australian Stock
Exchange was a normally distributed random variable with a mean of 5% and standard deviation of 10%.
The proportion of stocks that would have decreased in value, in that particular year, is closest to
A. 0.31
B. 0.5
C. 0.69
D. 0.05
E. 0.1
How do you get option A as the answer?
Note that it is the
percentage increase that is the normally distributed random variable, so we can make that X.
i.e.
)

(as this corresponds to 5%)

)
Then (on the ti-nspire) use normCdf [Menu] [5] [5] [2] (on mine anyway)
normCdf(-inf,0,0.05,.01) (don't write this calculator syntax on the exam, but this is MC anyway)
=0.3085=0.31 i.e. A
Also, when solving something like b2-4 < 0, how do you determine which way the inequality sign goes?
Thanks
You're values will be at the x-intercepts, so find them first.


Then the best way to go about it is to draw it out visually (just a little sketch).
So we draw out y=x
2-4, and look when it is below 0, that will be for

So that leaves us with
