A brick company found that 4% of its bricks were defective. If 400 bricks are checked at random, what is the?

A brick company found that 4% of its bricks were defective. If 400 bricks are checked at random, what is the probability that there are more than 22 defective bricks?








a. 0.2618





b. 0.1859





c. 0.1254





d. 0.0486





e. 0.0150





f. none of these

A brick company found that 4% of its bricks were defective. If 400 bricks are checked at random, what is the?
This probability can be approximated using a normal approximation. That is because





np = 400(0.04) = 16 %26gt;= 5 and


n(1-p) = 400(0.96) = 384 %26gt;= 5.





So, we will use a normal where





mean = np = 16 and


sd = sqrt{np(1-p)} = sqrt(400*0.04*0.96) = 3.92.





Since we are using the normal, then we will have to make a correction on the 22. Since we want to find the proportion above 22, but not including 22, then we will want to start at 22.5 and shade to the right on our normal. So now this is just a normal, so find z.





z = (22.5-16)/3.92


= 1.66





Using the standard normal table, you get the area to the right as 0.0485, so the answer is d.
Reply:f
Reply:B?

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