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7. An assembly line at a plant produces exactly 10000 widgets a day. Suppose that approximately...

7. An assembly line at a plant produces exactly 10000 widgets a day. Suppose that approximately 1 out of every 2000 fails a standards test and is thrown out. What is the probability that there will be 10 or more widgets thrown out on a given day?

8. There are 100 green balloons and 150 red balloons in a bag. Suppose we extract 10 balloons from the bag. (a) What is the exact probability that five of the balloons will be green? (b) Use Binomial Approximation to find the probability that exactly five of the balloons will be green. (c) Use Binomial Approximation to find the probability that no more than four of the balloons are green

Expert Solution

7.

Probability of failure, p = 1/2000

Number of trials, n = 10000

Then the number of widgets thrown out on a given day will follow Binomial distribution with parameter n = 10000 and 1/2000

Using Normal approximation to the binomial distribution with mean = np = 10000 * 1/2000 = 5 and standard deviation = = 2.2355

Probability that there will be 10 or more widgets thrown out on a given day = P(X 10)

= P(X > 9.5) (Using Continuity correction)

= P[Z > (9.5 - 5)/2.2355]

= P[Z > 2.01]

= 0.0222

8.

a.

Using Hypergeometric distribution, probability that exactly five of the balloons will be green

= (¹⁰⁰C₅ * ¹⁵⁰C₅ ) / ²⁵⁰C₁₀

= 0.2033745

b.

Probability of a green balloons = 100/250 = 0.4

Use Binomial Approximation, number of green balloons, X ~ Binomial(n = 10, p = 0.4)

Probability that exactly five of the balloons will be green = ¹⁰C₅ * 0.4⁵ * 0.6⁵ = 0.2006581

c.

Probability that no more than four of the balloons are green = P(X 4)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= ¹⁰C₀ * 0.4⁰ * 0.6¹⁰ + ¹⁰C₁ * 0.4¹ * 0.6⁹ + ¹⁰C₂ * 0.4² * 0.6⁸ + ¹⁰C₃ * 0.4³ * 0.6⁷ + ¹⁰C₄ * 0.4⁴ * 0.6⁶

= 0.006046618 + 0.040310784 + 0.120932352 + 0.214990848 + 0.250822656

= 0.6331033

milcah answered 3 months ago

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