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)?


Answer:
Given Data:

• Probability of success $(p)=0.1$
• Sample size $(n)=200$

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Solution:

The mean:

$\mu =n\times p$

$=200\times 0.1$

$=20$

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The standard deviation:

$\sigma =\sqrt{n\times p(1–p)}$

$=\sqrt{200\times 0.1\times (1–0.1)}$

$=4.243$

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A) The probability that $x$ is at most 30:

$P(X_{binomial}<30)=P(X_{normal}<30+0.5)$

$=P(\frac{X_{normal}–\mu }{\sigma }<\frac{30.5–20}{4.243})$

$=P(z<2.47)$

$\approx 0.9932$

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B) The probability that $x$ is less than 30:

$P(X_{binomial}<30)=P(X_{normal}<30+0.5)$

$=P(\frac{X_{normal}–\mu }{\sigma }<\frac{30.5–20}{4.243})$

$=P(z<2.47)$

$\approx 0.9932$

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C) The probability that $x$ is between 15 and 25 (inclusive):

$P(15<X_{binomial}<25)=P(15–0.5<X_{normal}<15+0.5)$

$=P(\frac{14.5–20}{4.243}<\frac{X_{normal}–\mu }{\sigma }<\frac{25.5–20}{4.243})$

$=P(-1.3<z<1.3)$

$=P(z<1.3)–P(z<-1.3)$

$=0.9032–0.0968$

$=0.8064$