The marks obtained in a certain examination follow normal distribution with mean 30 and standard deviation 10. If 1000 students appeared in the examinations, calculate the number of students scoring less than 33 marks.

618

The correct answer is Option (4) → 618

Let X denote the marks obtained in the examination.

Given $μ = 30, σ = 10$, then $Z =\frac{X-30}{10}$

$P(X <33) = P\left(Z <\frac{33-30}{10}\right)= P(Z < 0.3)$

$= F(0.3) = 0.6179$

∴ Number of students scoring less than 33 marks

= 1000 × 0.6179 = 617.9 i.e. 618.

Hence, 618 students scored less than 33 marks.