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tensa zangetsu [6.8K]
3 years ago
6

A lot of 25 skylight covers are received at your construction site, and before installation are subjected to an acceptance testi

ng procedure. The procedure consists of selecting 5 covers at random, without replacement, and testing each one using projectile to simulate hail stone impacts. If 2 or fewer of the 5 covers crack, the remaining ones are accepted. Otherwise the lot is rejected and sent back to the supplier. Assume a lot actually has four defective covers (impact strength is not adequate to withstand the test) among the 25. What is the exact probability of lot acceptance? Hint: Hypergeometric.
Mathematics
1 answer:
8090 [49]3 years ago
6 0

Answer:

If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.

Step-by-step explanation:

We have a population of N=25 skylight covers, were K=4 are defective.

We sample n=5 covers, and we will accept the lot if k=2 or fewer are defective.

We will use the hypergeometric distribution to model this probabilities.

First, to be accepted, the sample can have 2, 1 or 0 defective covers, so the probability of being accepted is:

P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)

The probability that there are k defective covers in the sample is:

P(k)=\dfrac{\dbinom{k}{k}\dbinom{N-k}{n-k}}{\dbinom{N}{n}}

Then, we can calculate the individual probabilities as:

P(k=0)=\dfrac{\dbinom{4}{0}\cdot \dbinom{25-4}{5-0}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{0}\cdot \dbinom{21}{5}}{\dbinom{25}{5}}\\\\\\P(k=0)=\dfrac{1\cdot 20349}{53130}=0.38

P(k=1)=\dfrac{\dbinom{4}{1}\cdot \dbinom{25-4}{5-1}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{1}\cdot \dbinom{21}{4}}{\dbinom{25}{5}}\\\\\\P(k=1)=\dfrac{4\cdot 5985}{53130}=0.45

P(k=2)=\dfrac{\dbinom{4}{2}\cdot \dbinom{25-4}{5-2}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{2}\cdot \dbinom{21}{3}}{\dbinom{25}{5}}\\\\\\P(k=2)=\dfrac{6\cdot 1330}{53130}=0.15

If we add this probabilities, we have:

P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)\\\\P(accepted)=0.38+0.45+0.15=0.98

If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.

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On average, people visit Thrillville 2 more times per year than Funland.

<h3><u>Averages</u></h3>

To determine how many more times per year, on average, do people visit Thrillville than Funland, the following calculation must be performed:

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Step-by-step explanation:

Given the vectors based on the number line as RS = 7y +3, ST = 5y +8, and RT = 83, the equation RS+ST = RT will be used to get the unknown.

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