CA Foundation · Quantitative Aptitude · Theoretical Distributions
The marks of students in a test are normally distributed with mean 60 and standard deviation 5. Given that the area under the standard normal curve between z = 0 and z = 1 is 0.3413, and between z = 0 and z = 2 is 0.4772, what proportion of students score between 55 and 70 marks?
The proportion is 0.8185. Marks of 55 and 70 correspond to z-values of -1 and 2. By symmetry the area from -1 to 0 equals 0.3413, and the area from 0 to 2 is 0.4772. Adding them gives 0.8185.
- A0.1359
- B0.8185Correct
- C0.7185
- D0.9544
Explanation
z for 55 is (55 - 60)/5 = -1 and z for 70 is (70 - 60)/5 = 2. The area between -1 and 2 is 0.3413 + 0.4772 = 0.8185 by symmetry. The value 0.1359 is wrongly obtained by subtracting the two areas.
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