CA Foundation · Quantitative Aptitude · Theoretical Distributions
The monthly electricity bills of households in a Nagpur colony are normally distributed with mean ₹1,200 and standard deviation ₹200. Using the standard normal table, P(0 < Z < 1) = 0.3413. What proportion of households have bills between ₹1,200 and ₹1,400?
The proportion is 0.3413. Converting ₹1,400 gives Z = (1,400 - 1,200)/200 = 1, and ₹1,200 gives Z = 0, so the area between the mean and one standard deviation above it is 0.3413.
- A0.1587
- B0.3413Correct
- C0.6826
- D0.8413
Explanation
Z for 1,400 = (1,400 - 1,200)/200 = 1. Z for 1,200 = 0. So the required probability is P(0 < Z < 1) = 0.3413. The option 0.6826 is wrong because it covers both sides of the mean (-1 to 1).
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