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 (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. (Round your answers to four decimal places.) (a) What is the (approximate) probability that X is at most 30? (b) What is the (approximate) probability that X is less than 30? (c) What is the approximate) probability that X is between 15 and 25 (inclusive)?

Explanation / Answer

We are given,

P = 0.10

n = 200

np = 200 *0.10

      = 20

nq = 200 *(1-0.10)

      = 180

Here both np and nq are greater than 5 so we use normal approximation to binomial.

Question a)

P (x <= 30)

We use correction factor and we get,

P (x < 30.5)

Mean = np = 20

Standard deviation = sqrt (npq)

                                   = sqrt (200*0.10*(1-0.10))

                                   = 4.2426

z = (x – Mean) / Standard deviation

= (30.5-20)/4.2426

=2.47

P (z<= 2.47)

= 0.9932              [By using z-table]

Answer: 0.9932

Question b)

P (x <30)

By using correction factor,

P (x <29.5)

z = (x – Mean) / Standard deviation

= (29.5-20)/4.2426

=2.24

P (z<2.24)

= 0.9875    [By using z-table]

Answer: 0.9875

Question c)

P (15 <= x <= 25)

By using correction factor we get,

P (14.4 <= x <= 25.5)

z1 = (x – Mean) / Standard deviation

= (14.5-20)/4.2426

= -1.30

z2 = (x – Mean) / Standard deviation

= (15.5-20)/4.2426

= 1.30

P (-1.30 <= z <= 1.30)

= P (z<= 1.30) – P (z<= 1.30)

= 0.9032 – 0.0968             [By using z-table]

=0.8064

Answer: 0.8064

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