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Vlad1618 [11]
3 years ago
15

A machine that manufactures automobile pistons is estimated to produce a defective piston 1% of the time. Suppose that this esti

mate is correct and that a random sample of 90 pistons produced by this machine is taken. Estimate the number of pistons in the sample that are defective by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.
Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

Mean: 0.9 pistons

Standard deviation: 0.944 pistons

Step-by-step explanation:

Either a piston is defective, or it is not. This means that for either piston, there can only be two outcomes. This means that we can solve this problem by the binomial distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and each trial can only have two outcomes, with probabilities \pi and 1 - \pi.

For an experiment with n repeated trials and a probability of success \pi, we have that:

The mean is: \mu_{x} = n*\pi

The standard deviation is: \sigma_{x} = \sqrt{n*\pi*(1-\pi)}

In this problem

There are 90 pistons, so n = 90.

There is a 1% probability that a piston is defective, so \pi = 0.01.

Estimate the number of pistons in the sample that are defective by giving the mean of the relevant distribution:

\mu_{x} = n*\pi = 90(0.01) = 0.9

Quantify the uncertainty of your estimate by giving the standard deviation of the distribution.

\sigma_{x} = \sqrt{n*\pi*(1-\pi)} = 0.944

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Delta makes 12-volt car batteries. These batteries are known to be normally
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Answer:

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the sample size n =12 -volt car batteries</em>

<em>Let  'X' be a Random variable in a normal distribution</em>

<em>Given that mean of the normal distribution = 45 months</em>

<em>Given that the Standard deviation of the normal distribution = 8months</em>

<u><em>Step(ii):-</em></u>

Let  X₁ = 3 years = 12 × 3 = 36 months

Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{36-45}{8} = -1.125

Let X₂ = 4 years  = 12 × 4 = 48 months

Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{48-45}{8} = 0.375

<u><em>Step(iii)</em></u>:-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = P(-1.125≤Z≤0.375)

                   = P(Z≤0.375) - P(Z≤-1.125)

                   = 0.5 +A(0.375) - (0.5-A(1.125)

                   = 0.5 + 0.1480 - (0.5 -0.3708)

                  = 0.1480 + 0.3708

                 = 0.5188

<u><em>Final answer:-</em></u>

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

<em />

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