Pessimist Singularitarian

#7
July 16th, 2013,
19:34
Originally Posted by

**karush**
(c) Given that $ \displaystyle P(X>105)=0.16$ (correct to $ \displaystyle 2$ significant figures), find $ \displaystyle P(d<X<105)$

so that is within $ \displaystyle 68\%$ within

$ \displaystyle 1$ standard deviation of the mean

or do just $ \displaystyle (2)0.16 = 0.32$

not sure??

I would write (for clarity):

$ \displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$

$ \displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$

$ \displaystyle 0.68=P(d<X<100)+0.34$

$ \displaystyle P(d<X<100)=0.34$

What do you find?