Gaussian exam mark

Problem

An exam mark follows $N(25, 16)$. What is the probability of a mark $\ge 30$? (Use the $68$-$95$-$99{,}7$ rule or the standard normal table.)

Solution

Here $\mu = 25$ and $\sigma^2 = 16$, so $\sigma = 4$. We standardise the value $30$: $z = \frac{30-\mu}{\sigma} = \frac{30-25}{4} = 1{,}25.$ From the standard normal table $\Phi(1{,}25) \approx 0{,}894$, hence $P(X\ge 30) = 1 - \Phi(1{,}25) \approx 1 - 0{,}894 = \boxed{0{,}106}.$ About $11\%$ of the students score above mark $30$. (With the $68$-$95$-$99{,}7$ rule alone one can only estimate: $30$ lies between $\mu+\sigma=29$ and $\mu+2\sigma=33$, so the probability is between $2{,}5\%$ and $16\%$.)

Links

Topics: Distribuzioni probabilita
Concepts: Deviazione standard · Distribuzione normale
Skills: Calcolo probabilita · Stimare
Exercise type: Problema probabilita