Suppose that 10% of all steel shafts produced by a certain process are nonconfor

Question

Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked. Consider a random sample of 200 shafts, and let X denote the number of shafts among these that are nonconforming and can be reworked. Find the following answers using the normal approximation to the binomial distribution:


a. Find the mean number of shafts and the standard deviation of the number of shafts that are nonconforming and can be reworked?


b. What is the probability that the number of shafts is between 18 and 22 (inclusive)?



c. What is the probability that the number of shafts is greater than 23?

Explanation / Answer

a)

mean =np =200*0.1 =20

std deviation =sqrt(np(1-p)) =4.243

b) probability that the number of shafts is between 18 and 22 (inclusive)

=P(18<=X<=22)=P((17.5-20)/4.243<Z<(22.5-20)/4.243)=P(-0.59<Z<0.59)=0.7224-0.2776 =0.4448

c)

probability that the number of shafts is greater than 23:

P(X>23) =1-P(X<=23)=1-P(Z<(23.5-20)/4.243)=1-P(Z<0.82)=1-0.7939 =0.2061

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