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adelina 88 [10]
3 years ago
10

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci

rcuits 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
Mathematics
1 answer:
mafiozo [28]3 years ago
7 0

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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3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

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F(v) = F'(x)

Differentiate F(x) to give:

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F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

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