The weekly sales (in thousands of euros), in a certain company are normally distributed with $\mu = 100$ and $\sigma^2 = 9$. What is the probability that the average of the sales, in the following 5 weeks, doesn't exceed 102? Assume the sales in the different weeks are independent.
I did:
$$P(X \le 102) = \phi(\frac{102-100}{3}) = 0.745373$$
Then I used the binomial distribution:
$$\binom{5}{5}(0.745373)^5(1-0.745373)^0 = 0.2300743932$$
Which is wrong. The right answer is 0.931888. Help?
What you didn't realize is that they want ON AVERAGE the sales not to exceed $102$. So you need to apply the $\sqrt{n}$ law, meaning that $3$ needs to be divided by $\sqrt{5}$ in the denominator. That gives you a $z$ score of $1.49$. Looking up in the $z$ table gives you the correct answer