Question

Students are usually tested on various subjects to determine the level of intelligence. In an intelligence examination administered to 2,000 students of the MBA programmes, the average score recorded by the Lecturer was 42 and deviation of the scores from their mean was 24. Assuming that the distribution of scores is normal, find:
i. The number of students whose score exceeds 50. (4 Marks)
ii. The number of students whose score is between 30 and 54 inclusive. (4 Marks)
iii. The score corresponding to the standardized normal value of 2. (4 Marks)
iv. What is the probability of a score which is less than 70? (4 Marks)
v. How many student would have a score of at least 66?(4 Marks)

Answer 1

i. The number of students whose score exceeds 50:

$$P(X>50)=P\bigg(z>\frac{50-42}{24}\bigg)=P(z>0.333)=\\ =0.5-P(0\leq z\leq0.333)=0.5-0.1293=0.3707,\\ X=0.3707\cdot2000=742.$$

ii. The number of students whose score is between 30 and 54 inclusive:

$$P(30\leq z\leq54)=P\bigg(\frac{30-42}{24}\leq z\leq\frac{54-42}{24}\bigg)=\\=P(0\leq z\leq0.5)+P(0\leq z\leq0.5)=\\ =P(-0.5\leq z\leq0.5)=\\ =0.1915+0.1915=0.383,\\ X=0.383\cdot2000=766.$$

iii. The score corresponding to the standardized normal value of 2:

$$z=\frac{X-\mu}{\sigma}=\frac{2-42}{24}=-1.667.$$

iv. What is the probability of a score which is less than 70?

$$P(-\infty< z\leq70)=P\bigg(\frac{-\infty-42}{24}\leq z<\frac{70-42}{24}\bigg)=\\ =P(-\infty\leq z\leq1.167)=\\ =0.8770.$$

v. How many student would have a score of at least 66?

$$P(X>66)=P\bigg(z>\frac{66-42}{24}\bigg)=P(z>1)=\\ =0.5-P(0\leq z\leq1)=0.5-0.3413=0.1587,\\ X=0.1587\cdot2000=318.$$