Question

The probability of the standard normal variable taking values between 0 and 1 is 0.3413, between 0 and 2 is 0.4772, and between 0 and 3 is 0.4987. The average of marks in an examination is 68 and the standard deviation is 10. The percentage of examinees getting less than 48 marks is:

• A. 2.28
• B. 10.31
• C. 47.72
• D. 52.78

Hint:

- Use the Z-score formula \( Z = \frac{X - \mu}{\sigma} \) to standardize a value to the standard normal distribution.
- The area under the standard normal curve gives the probability, which can be converted into percentages.

Answer 1

The Correct Option is A

Let \( X \) be the marks obtained by an examinee. We are given that the mean \( \mu = 68 \) and the standard deviation \( \sigma = 10 \). We need to find the percentage of examinees scoring less than 48 marks. First, we standardize the score using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Substituting the values: \[ Z = \frac{48 - 68}{10} = \frac{-20}{10} = -2 \] Now, we need to find \( P(Z<-2) \). From the standard normal distribution table, we know that: \[ P(Z<-2) = 0.0228 \] Thus, the percentage of examinees scoring less than 48 marks is: \[ 0.0228 \times 100 = 2.28% \] Hence, the correct answer is (A) 2.28.