How to approximate the expected value in this problem I was solving this probability problem and I don't know how to approximate the expected value.
Thanks in advance!
Problem definition:

The durability of a tire in a city of South Africa is a random variable with exponential distribution with parameter:
  
  
*
  
*0.29 if the temperature (in Farenheit) at the time of inflating the tire is greater than 70 degrees. 
  
*0.14 if the temperature (in Farenheit) at the time of inflating the tire is less or equal than 70 degrees. 
  
  
  The temperature in that city (in Farenheit) has a normal distribution $\mathcal{N}(68, 26)$.
102 tires are randomly chosen, all of them inflated in different days. Approximate the probability that the average of the durabilities of these tires is less than 6 years. 

 A: So formally, we know that $E(E(Y|X))=E(Y),$ where $Y$ is the durability of one tire and $X$ is the temperature, split as $X>70$ and $X\le70$.  What we are essentially going to do is take a weighted average: $$E(Y)=E(Y|X>70)\Pr(X>70)+E(Y|X\le70)\Pr(X\le70)$$
The mean of an exponential distribution with parameter $\lambda$ is $\frac{1}{\lambda}$, so, using this and a normal distribution to find the probabilities on the temperature $X$ (I'm assuming 26 is the variance, not the standard deviation, so this part might be incorrect, but the idea is the same), we get:
$$E(Y)=\frac{1}{0.29}\cdot0.347+\frac{1}{0.14}\cdot0.653\\
=5.859$$
Now we can use a similar method to get the variance on $Y$, this time the formula being:
$$Var(Y)=E(Var(Y|X))+Var(E(Y|X))$$
Here, the variance of an exponential distribution is $\frac{1}{\lambda^2}$, so this becomes:
$$E(Var(Y|X))=Var(Y|X>70)\Pr(X>70)+Var(Y|X\le70)\Pr(X\le70)\\
=\frac{1}{0.29^2}\cdot0.347+\frac{1}{0.14^2}\cdot0.653\\
=37.425$$
and
$$Var(E(Y|X))=[E(Y|X>70)-E(Y)]^2\Pr(X>70)+\\ [E(Y|X\le70)-E(Y)]^2\Pr(X\le70)\\
=[\frac{1}{0.29}-5.859]^2\cdot0.347+[\frac{1}{0.14}-5.859]^2\cdot0.653\\
=3.095$$
Putting these together, we get that the variance on the durability for a single tire is about $40.52$
Now, 102 is a reasonably large number of tires, so it is reasonable to approximate the average durability of the tires using a normal distribution.  The mean will still be $5.859$, and the variance will then be $\frac{40.52}{102}$.  Hopefully this helps!
