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Vikki [24]
3 years ago
9

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the en

tire batch is accepted if every item in the sample is okay. The ABC Electronics Company has just manufactured 2100 write-rewrite CDs, and 60 are defective. If 3 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

0.9167 = 91.67% probability that the entire batch will be accepted

Step-by-step explanation:

The CDs are chosen without replacement, which means that the hypergeometric distribution is used.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

2100 write-rewrite CDs, and 60 are defective.

This means, respectively, that N = 2100, k = 60

3 of these CDs are randomly selected for testing

This means that n = 3

What is the probability that the entire batch will be accepted?

Probability of none defective, so P(X = 0).

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 0) = h(0,2100,3,60) = \frac{C_{60,0}*C_{2040,3}}{C_{2100,3}} = 0.9167

0.9167 = 91.67% probability that the entire batch will be accepted

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Answer:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low\leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can consider a value to be significantly low if we have that the z  score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is  higher tor equal to 2.

For this case we have the mean and the deviation given:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low \leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1

6 0
3 years ago
During a period of 80 minutes, a music station played 8 minutes of commercials. What is the ratio of music they played to commer
igomit [66]

Answer:

Step-by-step explanation:

80 minutes.......8 minutes played commercials....that means (80 - 8) = 72 minutes played music

so the ratio of music to commercials is : 72/8 which reduces to 9/1...so the ratio is 9:1 <=== ur answer

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8 0
3 years ago
Need help adding these<br><br> √54 + √24
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Based on a study of population projections for 2000 to​ 2050, the projected population of a group of people​ (in millions) can b
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Answer:

(a) 0.107 million per year

(b) 0.114 million per year

Step-by-step explanation:

A(t) = 11.19(1.009)^t

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A\prime(t) = \dfrac{A}{t}=11.19\cdot1.009^t\ln1.009

In the year 2014, t=14.

A\prime(14) =11.19\cdot1.009^14\ln1.009=0.114

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