One person is randomly selected from a population whose mean is 45 and the standard deviation is 9. What is the probability that the person’s score is greater than 48?

Answer :

Given :

The population mean $(μ) = 45$

The population standard deviation $(σ) = 9$

Solution :

→ The probability that the person’s score is greater than 48 :

$$\therefore \text{P}(x > 48) = \text{P}\left(\frac{x – \mu}{\sigma} > \frac{48 – 45}{9}\right)$$ $$= \text{P}(z > 0.33)$$ $$= 0.3707$$