# Mean & Standard Deviation

• Jun 15th 2009, 02:47 AM
TS5
Mean & Standard Deviation
A steel manufacturer finds that the total quantity of raw materials used by his company each week is normally distributed with a mean of 1000 tonnes and a standard deviation of 200 tonnes. Calculate an estimate of the maximum quantity of raw materials the company is likely to use per week.

I thought that it is 1200 because mean is the average and so with the standard deviation it would be 1000+200=1200. But, the answer accepted '1400 or 1600 tonne', how is this calculated? :confused:
• Jun 15th 2009, 04:42 AM
mr fantastic
Quote:

Originally Posted by TS5
A steel manufacturer finds that the total quantity of raw materials used by his company each week is normally distributed with a mean of 1000 tonnes and a standard deviation of 200 tonnes. Calculate an estimate of the maximum quantity of raw materials the company is likely to use per week.

I thought that it is 1200 because mean is the average and so with the standard deviation it would be 1000+200=1200. But, the answer accepted '1400 or 1600 tonne', how is this calculated? :confused:

Obviously they decided to accept $\mu + 2 \sigma$ and $\mu + 3 \sigma$ as well.

The question is actually too vague to give a definitive answer to. I'd suggest the justification of the answer is more important than the actual answer itself.