CA Foundation · Quantitative Aptitude · Theoretical Distributions
The monthly electricity bills of households in a Jaipur colony are normally distributed with mean Rs 2,000 and standard deviation Rs 400. Given P(0 < Z < 1.25) = 0.3944, what is the probability that a household's bill exceeds Rs 2,500?
The probability is 0.1056. Converting Rs 2,500 gives Z = (2500 - 2000)/400 = 1.25. The right tail beyond it is 0.5 minus the table area 0.3944, which equals 0.1056.
- A0.3944
- B0.8944
- C0.1056Correct
- D0.2056
Explanation
Z = (2500 - 2000)/400 = 1.25. P(X > 2500) = P(Z > 1.25) = 0.5 - 0.3944 = 0.1056. Option 0.3944 is only the area between mean and 2500, and 0.8944 is P(Z < 1.25), the opposite tail.
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