MHB Probability that the total annual profit is not more than 800000

[mathmari]

Hey!

An insurance office has $2500$ contracts with mean annual profit (per contract) $\mu=330$ and standard deviation $\sigma=540$. Calculate the probability that the total annual profit is not more than $800000$.

I have one the following:

The annual total profit should be not more than $800000$, that means that per contract it shoulebe not more than $\frac{800000}{2500}=320$.

So $$Z=\frac{X-\mu}{\sigma}=\frac{320-330}{540}\approx -1.31 \\ P(Z\leq -1.31)=0.0968$$ Is that the correct probability?

 

[I like Serena]

Hey mathmari!

Let $n=2500$ be the number of contracts,
Let $X_i$ be the annual profit of contract $i$.
Let $X=\sum\limits_{i=1}^{n} X_i$ the annual profit.

So we want to know $P(X\not >800k)=P(X\le 800k)$.

We assume that the $X_i$ are independent and have distribution $N(\mu, \sigma^2)$.
Can we tell what the distribution of $X$ is?

 

[mathmari]

Let $n=2500$ be the number of contracts,
Let $X_i$ be the annual profit of contract $i$.
Let $X=\sum\limits_{i=1}^{n} X_i$ the annual profit.

So we want to know $P(X\not >800k)=P(X\le 800k)$.

We assume that the $X_i$ are independent and have distribution $N(\mu, \sigma^2)$.
Can we tell what the distribution of $X$ is?

Isn't $X$ also normal with $N(2500 \mu, 2500 \sigma^2)$ ?

 

[I like Serena]

Isn't $X$ also normal with $N(2500 \mu, 2500 \sigma^2)$ ?

Yep. (Nod)

Can we calculate $P(X\le 800k)$ now?
I think it comes out slightly different than what you wrote before.

 

[mathmari]

Yep. (Nod)

Can we calculate $P(X\le 800k)$ now?
I think it comes out slightly different than what you wrote before.

It is $$P\left (\frac{\sum_{i=1}^{2500}X_i-2500\mu}{2500\sigma}\leq 8000000\right )$$ or not?

 

[mathmari]

We have to apply central limit theorem, or not? :unsure:

 

[I like Serena]

We have to apply central limit theorem, or not?

No need.
We already know that the sum of independent random variables with identical normal distributions is again a normal distribution.
So we don't need the $n=2500$ to be sufficiently large.

It is $$P\left (\frac{\sum_{i=1}^{2500}X_i-2500\mu}{2500\sigma}\leq 8000000\right )$$ or not?

As you wrote before, the variance is $2500\sigma^2$.
So the standard deviation cannot be $2500\sigma$, can it? (Shake)

Btw, we need parentheses in the numerator.
Otherwise we have $\sum_{i=1}^{2500}X_i-2500\mu = (\sum_{i=1}^{2500}X_i)-2500\mu \ne \sum_{i=1}^{2500}(X_i-2500\mu)$.