SOLUTION: The weights (in pounds) of metal discarded in one week by households are normally distributed with a mean of 2.22 lb. and standard deviation of 1.09 lb. a) If one household is r

Question 1147535: The weights (in pounds) of metal discarded in one week by households are normally distributed with a mean of 2.22 lb. and standard deviation of 1.09 lb.
a) If one household is randomly selected, find the probability that it discards more than 2 lb of metal in a week.
b) Find a weight x so that the weight of metal discarded by 70% of the houses is above x.

Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
a) If one household is randomly selected, find the probability that it discards more than 2 lb of metal in a week.

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.

= = -0.22/1.09 = -0.20

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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b) Find a weight x so that the weight of metal discarded by 70% of the houses is above x.

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.)

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:

= -0.52

= -0.52

x - 2.22 = -0.52(1.09)

x - 2.22 = -0.5668

x = 1.6532

So, 70% of the households discard more than 1.6532 lbs. of metal.
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