Question

The final exam scores in a statistics class were normally distributed with a mean of 70 and a standard deviation of five. What is the probability that a student scored less than 55% on the exam?

Answer 1

Solution

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .………..........................................…….......…...…(1)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables......... (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) …........................................(2b)

Now, to work out the solution,

Let X = final exam scores (%) in a statistics class.

We are given: X ~ N(70, 5)............................................................................................................................................ (3)

Probability that a student scored less than 55% on the exam

= P(X < 55)

= P[Z < {(55 - 70)/5}] [vide (1) and (3)]

= P(Z < - 3)

= 0.0013 [vide (2b)] Answer

DONE

[Going beyond,

The above probability could have been found without using (2a) or (2b), by applying the following rule:

Empirical rule, also known as 68 – 95 – 99.7 percent rule: applicable to symmetric (bell-shaped) distributions

P{(µ - σ) ≤ X ≤ (µ + σ)} = 0.68; ..........……………………………………………….(4a)

P{(µ - 2σ) ≤ X ≤ (µ + 2σ)} = 0.95; ...............................……….....................…….(4b)

P{(µ - 3σ) ≤ X ≤ (µ + 3σ)} = 0.9973 ......................................…………………….(4c)

i.e., Mean ± 1 Standard Deviation holds 68% of the observati………………….(4d)

Mean ± 2 Standard Deviations holds 95% of the observations ………….…….(4e)

and Mean ± 3 Standard Deviations holds 99.73% of the observations. ……….(4f).

So, vide (4c) or (4f), probability > 3 or < - 3 = 0.0027 and so P(< - 3) = 0.0027/2 = 0.0013]