Mathematics, 23.04.2020 04:16 gleatons429

Of all the registered automobiles in a city, 8% fail the emissions test. Nine automobiles are selected at random to undergo an emissions test. Round the answers to four decimal places.

a. Find the probability that exactly three of them fail the test. The probability that exactly three of them fail the test is .

b. Find the probability that fewer than three of them fail the test. The probability that fewer than three of them fail the test is .

c. Find the probability that more than two of them fail the test. The probability that more than two of them fail the test is .

d. Would it be unusual for none of them to fail the test?

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Answer from: antoniocapetillo80

a) The probability that exactly three of them fail the test is 2.61%.

b) The probability that fewer than three of them fail the test is 97.02%.

c) The probability that more than two of them fail the test is 2.98%.

d) It would not be unusual for none of them to fail the test

Step-by-step explanation:

For each automobile, there are only two possible outcomes. Either it fails the test, or it does not. The probability of an automobile faiiling the test is independent of other automobiles. So we use the binomial probability distribution 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

8% fail the emissions test.

This means that p = 0.08

Nine automobiles.

This means that n = 9

a. Find the probability that exactly three of them fail the test.

This is P(X = 3)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{9,3}.(0.08)^{3}.(0.92)^{6} = 0.0261$

The probability that exactly three of them fail the test is 2.61%.

b. Find the probability that fewer than three of them fail the test.

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.08)^{0}.(0.92)^{9} = 0.4722$

$P(X = 1) = C_{9,1}.(0.08)^{1}.(0.92)^{8} = 0.3695$

$P(X = 2) = C_{9,2}.(0.08)^{2}.(0.92)^{7} = 0.1285$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4722 + 0.3695 + 0.1285 = 0.9702

The probability that fewer than three of them fail the test is 97.02%.

c. Find the probability that more than two of them fail the test.

Either fewer than three(two or less) fail, or more than two do. The sum of the probabilities of these events is 100%. So

97.02 + p = 100

p = 2.98

The probability that more than two of them fail the test is 2.98%.

d. Would it be unusual for none of them to fail the test?

More than 2.5 standard deviations from the mean is unusual.

E(X) = np = 9*0.08 = 0.72

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{9*0.08*0.72} = 0.81$

0 > 0.72 - 2*0.81

It would not be unusual for none of them to fail the test

Answer from: Quest

It would be 18 cm

Answer from: Quest

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.