An automobile manufacturer has discovered that 20% of all the transmissions it installed in a particular style of truck are defective after a year of use. It has contacted customers and asked them to return to their dealer after a year to have their transmission checked. The Friendly Auto Mart sold seven of these trucks and has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission

Answers

Answer 1

Answer:

0.148 = 14.8% probability that they will need to order at least one more new transmission

Step-by-step explanation:

For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

20% of all the transmissions it installed in a particular style of truck are defective after a year of use.

This means that [tex]p = 0.2[/tex]

Sold seven trucks:

This means that [tex]n = 7[/tex]

It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?

This is the probability that at least 3 are defective, that is:

[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]

In which

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097[/tex]

[tex]P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670[/tex]

[tex]P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753[/tex]

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852[/tex]

[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148[/tex]

0.148 = 14.8% probability that they will need to order at least one more new transmission


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Answers


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Answer:

C, the graph is increasing when x is between -1 and 2.

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Answer:

1. Yes

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Answers

Answer:

A sample of 139 should be taken.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

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How large a sample should be taken if the desired margin of error is $500?

A sample of n is needed.

n is found when M = 500. So

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[tex]500\sqrt{n} = 1.96*3000[/tex]

Dividing both sides by 500

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A sample of 139 should be taken.

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