The distribution of weights of items produced by a manufacturing process can be approximated by a normal distribution with a mean of 90 grams and a standard deviation of 1 gram. Using the empirical rule, what percentage of the items will either weigh less than 88 grams or more than 92 grams?

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Answer 1 · 2015-03-05T03:46:18

The empirical rule: $34 - 13.5 - 2.5$ $34-13.5-2.5$

Which means that on each side of the mean $μ$ $mu$ 34% is within 1 standard deviation $σ$ $sigma$ (so 68% is within one $σ$ $sigma$ ).

On each side 13.5% is between $σ$ $sigma$ and $2 σ$ $2sigma$

If we add up $95 %$ $95%$ is between $μ - 2 σ$ $mu-2sigma$ and $μ + 2 σ$ $mu+2sigma$

Since $σ = 1 g \to 2 σ = 2 g$ $sigma=1g->2sigma=2g$

Answer: $100 % - 95 % = 5 %$ $100%-95%=5%$
2.5% weigh less than 88 g and 2.5% weigh more than 92 g.

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