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The HR manager needs a budget for recruitment costs for next year. FRom previous records they know that the cost of a single recruitment is normally distributed with a mean of 1,200 and a standard deviation 200.

What is the probability that the total of the next 10 recruitment will exceed 13,000?

What i did was 13000-1200 ÷ 200/root10 and got 186.57 however in the solutions it said the first step is to 13000/10 which i dont understand?

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Assuming each recruitment is independent then the sum of the next 10 recruitments is normally distributed with mean $10*1200$ and variance $10*200^2$

You want to calculate the probability of this distribution exceeding 13000.

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  • $\begingroup$ How do i put that into the equation? $\endgroup$ – Riv Sep 7 '15 at 22:23
  • $\begingroup$ Use the formula $Z=(X-\mu)/\sigma$ to standardise the normal variable. $\endgroup$ – WMycroft Sep 7 '15 at 22:31

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