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zheka24 [161]
3 years ago
12

Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take

a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part. The mean of this situation is?
Mathematics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components = \frac{68}{400} = 0.17

let q = probability of a non-defective computer components = \frac{332}{400} = 0.83

Given random sample n = 25

we will find mean value in binomial distribution

The mean of binomial distribution = np

here 'n' is sample size and 'p' is defective components

mean (μ) = 25 X 0.17 = 4.25

<u>Conclusion</u>:-

mean (μ) =  4.25

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Answer:

(x + 2) is a factor of f(x)

Step-by-step explanation:

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Could somebody help me with this question please?
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The members of a high school band asked a number of students whether they would like blue, gold, or both for the uniforms for th
MaRussiya [10]

The values of a and b are a = 33% and b = 43% ⇒ last answer

Step-by-step explanation:

The venn-diagram contains:

  • A circle labeled blue 32
  • A circle labeled gold 25
  • The two circles are overlap with label 12 on the overlap

The table:

→ Band  :  blue  :  N.b  : total

→ gold   :  16%   :  a      :  49%

→ N.g     :  b       :  8%   :  51%

→ Total  :  59%  :  41%  :  100%

We need to find the values of a and b

From the table

16% represents the percentage of blue and gold uniforms (2nd row with 2nd column)

From venn-diagram

The common part of blue and gold label with 12 (overlap of two circles)

∴ 16% of the total number of students = 12

∵ 16% × x = 12

∴ \frac{16}{100} × x = 12

∴ 0.16 x = 12

- Divide both sides by 0.16

∴ x = 75

∵ x represents the total number of students

∴ The total number of students were asked is 75

From the table

b represents the percentage of blue but not gold uniforms (3rd row and 2nd column)

From venn-diagram

The part that has blue and not gold labeled with 32 (the part of the blue circle not in the gold circle)

∴ 32 of the total number of students like the blue but not gold

- Divide 32 by the total 75 , then multiply the quotient by 100%

   to find the percentage of blue but not gold

∴ b = \frac{32}{75} × 100% = 42.6666%

∴ b ≅ 43%

From the table

a represents the percentage of gold nut not blue uniforms (2nd row and 3rd column)

From venn-diagram

The part that has gold and not blue labeled with 25 (the part of the gold circle not in the blue circle)

∴ 25 of the total number of students like the gold but not blue

- Divide 25 by the total 75 , then multiply the quotient by 100%

   to find the percentage of gold but not blue

∴ a = \frac{25}{75} × 100% = 33.3333%

∴ a ≅ 33%

The values of a and b are a = 33% and b = 43%

To check your answer complete the table you will find the total of it is 100% ( for the columns 16% + 43% = 59% , 33% + 8% = 41% , then 59% + 41% = 100%, for the rows 16% + 33% = 49% , 43% + 8% = 51% , then 49% + 51% = 100%)

Learn more:

You can learn more about the probability in brainly.com/question/3756853

#LearnwithBrainly

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