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31 December, 07:59

The number of unbroken charcoal briquets in a 20-pound bag filled at the factory follows a normal distribution with a mean of 450 briquets and a standard deviation of 20 briquets. the company expects that a certain number of the bags will be underfilled, so the company will replace for free the 5% of bags that have too few briquets. what is the minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free?

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  1. 31 December, 10:10
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    The interval of filled bags is at confidence interval of 95%.

    At 95% confidence interval, Z=1.96 (from Z tables).

    But,

    Z = (x-mean) / sd = > 1.96 = (x-450) / 20 = > x (upper limit) = 20*1.96+450489 bags

    Lower limit or minimum number bags to avoid replace = 450-1.96*20 = 411 bags
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