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The monthly salary x in a big organization is normally distributed with mean Rs. 3000 & Standard deviation Rs.250.

What should be the minimum salary of a worker in this organization so that he belongs to top 5% of workers.

Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

1 Answer
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Ley 'y' be the minimum salary of the top 5% worker

$ Z_1 = \frac{y - \mu}{\sigma} = \frac{y - 3000}{250} $

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P(X $\geq$ y) = 0.05

$ \implies P(Z \geq Z_1) = 0.05 \\ \therefore P(0 /leq Z \leq Z_1) = 0.45 \\ \therefore Z_1 = 1.65 \\ \therefore …

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