Question

The distribution of statistics marks of some sandwich students was found to be normal with a mean within a range of 61- 65 and a standard deviation of within a range of 7.8 – 8.6 (1dp). It was found out that there were 348 students who took the course. With an assumed mean and standard deviation, answer the questions that follow: a. If the minimum mark to qualify for an interview was 43. What is the probability that a student selected at random qualified for the interview and what percentage did not qualify? (5marks) b. If the cut-off mark for selection is 48, how many and what percentage of the students were selected? c. How many students scored a mark greater than 54 but less than 78? d. How many students from the group scored a mark greater than 68 but less than 80?

Answer 1

Let the Mean be 63 marks and standard deviation be 8.2 marks

Let X denote the marks obtained by a randomly selected student

n = 348

(a) Qualifying marks = 43

Probability that a student selected at random qualified for the interview = P(X ≥ 43)

= P{Z ≥ (43 - 63)/8.2}

= P(Z > -2.439) = 0.9927

And 0.73% did not qualify

(b) Qualifying marks = 48

Percentage of students that were selected = P(X ≥ 48)

= P{Z ≥ (48 - 63)/8.2}

= P(Z ≥ -1.829)

= 0.9663 = 96.63%

Number of students qualified = 348*0.9663 = 336.27 ≈ 336

(c) Probability that a student scores greater than 54 but less than 78 = P(54 < X < 78)

= P{(54 - 63)/8.2 < Z < (78 - 63)/8.2}

= P(-1.10 < Z < 1.83)

= 0.8307

Thus, the required number of students = 0.8307*348 = 289.08 ≈ 289

(d)

Probability that a student scores greater than 68 but less than 80 = P(68 < X < 80)

= P{(68 - 63)/8.2 < Z < (80 - 63)/8.2}

= P(0.61 < Z < 2.07)

= 0.2517

Thus, the required number of students = 0.2517*348 = 87.6 ≈ 88