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The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 12 defectives.(a) Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.< p\l>(b) Calculate a 95% upper confidence bound on the fraction of defective circuits. Round the answer to 4 decimal places

Answers

Answer 1

Answer:

Answer:

(a) 0.0178 <= p <= 0.0622

(b) p <= 0.0586

Step-by-step explanation:

We have that the sample proportion is:

p = 12/300 = 0.04

(to)

For 95% confidence interval alpha = 0.05, so critical value of z will be 1.96

Therefore, we have that the interval would be:

p + - z * (p * (1-p) / n) ^ (1/2)

replacing we have:

0.04 + - 1.96 * (0.04 * (1-0.04) / 300) ^ (1/2)

0.04 + - 0.022

Therefore the interval would be:

0.04 - 0.022 <= p <= 0.04 + 0.022

0.0178 <= p <= 0.0622

(b)

For upper bounf z-critical value for 95% confidence interval is 1.645, so upper bound is:

p + z * (p * (1-p) / n) ^ (1/2)

replacing:

0.04 + 1.645 * (0.04 * (1-0.04) / 300) ^ (1/2)

0.04 + 0.0186 = 0.0586

p <= 0.0586

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Step-by-step explanation:

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Two teams A and B play a series of games until one team wins four games. We assume that the games are played independently and that the probability that A wins any game is p. What is the probability that the series lasts exactly five games?

Answers

Answer:

4pq(p³+q³)

Step-by-step explanation:

Exactly 5 game would be when 4 wins and 1 loss of a particular person

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On Monday, there were 827 people at the amusement park. On Tuesday, the weather was rainy and only 686 people were at the park. Write an expression to estimate the difference of the number of people atthe park on Monday and Tuesday by rounding each value to the nearest
hundred.