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CA Foundation · Quantitative Aptitude · Theoretical Distributions

The monthly electricity bills of households in a Jaipur colony are normally distributed with mean Rs 1,800 and standard deviation Rs 300. Given that the area under the standard normal curve between Z = 0 and Z = 1 is 0.3413, what proportion of households have bills above Rs 2,100?

The proportion is 0.1587. A bill of Rs 2,100 corresponds to Z = 1, and the area to the right of Z = 1 is the upper tail, equal to 0.5 minus 0.3413, which gives 0.1587.

  1. A0.3413
  2. B0.6826
  3. C0.1587Correct
  4. D0.8413

Explanation

Z = (2100 - 1800)/300 = 1. P(Z > 1) = 0.5 - 0.3413 = 0.1587. Option 0.3413 is wrong because it is the area between the mean and Z = 1, not the tail. Option 0.8413 is P(Z < 1).

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