The average test marks in a particular class is 79. The standard deviation is 5. Marks are normally distributed. We have to find how many students, in a class of 200, did not receive marks between 75 and 82.

Given: Mean (μ) = 79
Standard deviation (σ) = 5
Class size = 200
Given probabilities (for standard normal variable Z):
P(0 ≤ Z ≤ 0.7) = 0.2580
P(0 ≤ Z ≤ 0.8) = 0.2880
P(0 ≤ Z ≤ 0.6) = 0.2257

Step 1: Convert the raw scores to Z-scores.

For X = 75: Z1 = (75 − 79) / 5
Z1 = −4 / 5
Z1 = −0.8

For X = 82: Z2 = (82 − 79) / 5
Z2 = 3 / 5
Z2 = 0.6

So, we want the probability that marks are not between 75 and 82, i.e.
P(X < 75 or X > 82).
First we will find P(75 ≤ X ≤ 82), then subtract from 1.

Step 2: Find P(75 ≤ X ≤ 79) and P(79 ≤ X ≤ 82).

Because the normal distribution is symmetric about the mean:

P(75 ≤ X ≤ 79) = P(−0.8 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 0.8)
From the given values: P(0 ≤ Z ≤ 0.8) = 0.2880

P(79 ≤ X ≤ 82) = P(0 ≤ Z ≤ 0.6)
From the given values: P(0 ≤ Z ≤ 0.6) = 0.2257

Step 3: Probability of marks lying between 75 and 82.

P(75 ≤ X ≤ 82) = P(75 ≤ X ≤ 79) + P(79 ≤ X ≤ 82)
P(75 ≤ X ≤ 82) = 0.2880 + 0.2257
P(75 ≤ X ≤ 82) = 0.5137

Step 4: Probability of marks lying outside 75 and 82.

P(X < 75 or X > 82) = 1 − P(75 ≤ X ≤ 82)
P(X < 75 or X > 82) = 1 − 0.5137
P(X < 75 or X > 82) = 0.4863

Step 5: Convert probability into number of students.

Number of students not getting marks between 75 and 82
= 0.4863 × 200
= 97.26 ≈ 97 students

Final Answer:
Approximately 97 students in the class of 200 did not receive marks between 75 and 82. 

👉 Important questions of Statistical Analysis for Business

1. What do you understand by Probability Distribution? Explain the characteristics and applications of Normal Distribution.

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The average test marks in a particular class is 79.