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CA Foundation · Quantitative Aptitude · Theoretical Distributions

The daily sales of a Kolkata shop are normally distributed. It is known that 6.68% of days have sales above Rs 26,000 and the mean is Rs 20,000. Given that P(0 < Z < 1.5) = 0.4332, what is the standard deviation of daily sales?

The standard deviation is Rs 4,000. A 6.68% upper tail corresponds to Z = 1.5, so Rs 6,000 above the mean equals 1.5 standard deviations, giving 6,000 divided by 1.5, which is Rs 4,000.

  1. ARs 6,000
  2. BRs 4,000Correct
  3. CRs 3,000
  4. DRs 2,000

Explanation

Upper tail 0.0668 means area from 0 to Z is 0.5 - 0.0668 = 0.4332, so Z = 1.5. Then (26,000 - 20,000)/σ = 1.5, giving σ = 6,000/1.5 = 4,000. Check: 20,000 + 1.5(4,000) = 26,000. Rs 6,000 wrongly uses Z = 1.

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