Question 1147535
<i>a) If one household is randomly selected, find the probability that it discards more than 2 lb of metal in a week.</i>
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You want to find what the probability is that this household discards LESS than 2 lbs. of metal per week, then subtract that result from 1.
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{{{(2 - mean)/SD}}} = {{{(2 - 2.22)/1.09}}} = -0.22/1.09 = -0.20
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Look up -0.20 on a z-table and we get a result of 0.4207.  This is the probability a household discards LESS than 2 lbs. of metal per week.  Therefore, the probability a household discards MORE than 2 lbs. of metal per week is: 1 - 0.4207...or 0.5793.
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<i>b) Find a weight x so that the weight of metal discarded by 70% of the houses is above x.</i>
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Again, you want to look for the inverse...the weight x so that the weight of metal discarded by 30% of the households is BELOW x.  (The weight where metal is discarded by 30% of the households BELOW x is the exact same number as the weight where metal is discarded by 70% of the households ABOVE x.)
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So, we are looking for the weight x so that the weight of metal discarded by 30% of the households is BELOW x.  This means we want to go to a z-table and find the z-score that most closely corresponds to 0.3000.  The z-score where this is closest is -0.52.  So, we set up the equation this way:
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{{{(x - mean)/SD}}} = -0.52
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{{{(x - 2.22)/1.09}}} = -0.52
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x - 2.22 = -0.52(1.09)
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x - 2.22 = -0.5668
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x = 1.6532
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So, 70% of the households discard more than 1.6532 lbs. of metal.