# Normal Distribution – Applying Central Limit Theorem to Sum of 5 Normal Distributions

central limit theoremnormal distributionself-study

what would be a correct solution to following problem:

The medical costs per patient are normally distributed with a mean equal to 7 euro and a variance equal to 49. Compute the probability that the medical costs for five patients are more than 21 euro.

a) Linear transformation
$$X \sim N(7, 49),\hspace{.1cm} X \sim \text{med. cost for one patient}$$
$$Y = 5X, \hspace{.1cm} Y \sim \text{med. cost for five patients}$$
then $$E(Y) = 5\cdot7 = 35$$, and $$Var(Y) = 5^2 \cdot 49 = 1225$$

so $$P(Y>21) = P\left(Z> \frac{21-35}{\sqrt{5^2 \cdot 49}}\right) = P(Z >-0.4) = \Phi(0.4) = 0.66$$

b) CLT

$$S_5 = \sum_{i = 1}^{5} x_i$$

$$P(S_5> 21) = P\left(Z >\frac{21-35}{\sqrt{5 \cdot 49}}\right) = P(Z > -0.89) = 0.81$$

This question should really have a self study tag and even if it has not, the answer cannot be complete.

Now from Wikipedia we can learn, how sums of normally distributed values are distributed: https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances

In your above attempt at solving this you computed the new mean as $$5\times 7$$ and the new variance as $$5^2\times 49$$. Now which of these computations does what the Wikipedia tells you to do and which one does not?

Edit after @JakubL 'S comment under this answer:

You are correct now. If you use R, you can easily simulate that like this:

> n <- 1000000
> cost <- replicate(n, sum(rnorm(n = 5, mean = 7, sd = sqrt(49))))
> sum(cost > 21)/n
[1] 0.814186


And if we use $$Y$$ follows $$N(5\times 7, 5\times 49)$$ the result is the same:

> pnorm(21, 5*7, sqrt(5*49), lower.tail = FALSE)
[1] 0.8144533