Answer:
See proof below
Step-by-step explanation:
We have to verify that if we substitute in the equation
the equality is true.
Let's substitute first in the right hand side:
Now we use the distributive laws. Also, note that (this also works when the power is n-2).
then the sequence solves the recurrence relation.
Given mean = 0 C and standard deviation = 1.00
To find probability that a random selected thermometer read less than 0.53, we need to find z-value corresponding to 0.53 first.
z=
So, P(x<0.53) = P(z<0.53) = 0.701944
Similarly P(x>-1.11)=P(z>-1.11) = 1-P(z<-1.11) = 0.8665
For finding probability for in between values, we have to subtract smaller one from larger one.
P(1.00<x<2.25) = P(1.00<z<2.25) = P(z<2.25)- P(z<1.00) = 0.9878 - 0.8413 = 0.1465
P(x>1.71) = P(z>1.71) = 1-P(z<1.71) = 1-0.9564 = 0.0436
P(x<-0.23 or x>0.23) = P(z<-0.23 or z>0.23) =P(z<-0.23)+P(z>0.23) = 0.409+0.409 = 0.918