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Luba_88 [7]
3 years ago
12

One-hour carbon monoxide concentrations in air samples from a large city average 12 ppm (parts per million) with standard deviat

ion 9 ppm.
Required:
a. Do you think that carbon monoxide concentrations in air samples from this city are normally distributed.
b. Find the probability that the average concentration in 100 randomly selected samples will exceed 14 ppm.
Mathematics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

0.0132

Step-by-step explanation:

No

Carbon monoxide concentrations sample in air from a large city are not normally distributed. This is often so because of the time of the day and that of the night. There exists different concentrations during the day, with respect to during the night. The companies and factories work during the day, so does most vehicles. This means that there tend to be a higher concentration during the day, than there will be during the night.

b)

Assuming that Y' is approximately normal, we can say that

P(Y' > 14) = P(Z > [√100(14 - 12)/9]) and this is equal to

P(Z > 2.22) = 0.013

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Answer:

a. <u>Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.</u>

b. <u>The probability is 0.2029 or 20.29%</u>

c. <u>The probability is 0.2029 or 20.29%</u>

d. <u>The probability is 0.1671 or 16.71%</u>

e. <u>The probability is 0.9975 or 99.75%</u>

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<u>a. Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.</u>

b. Let's use the binomial distribution table, this way:

Binomial distribution (n=10, p=0.697)

 f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0000 0.0000

1 0.0002 0.0002

2 0.0016 0.0017

3 0.0095 0.0112

4 0.0384 0.0496

5 0.1059 0.1555

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10 0.0271 1.0000

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c. If 69.7% of 18-20 years old consumed alcoholic beverages in 2008, therefore, 30.3% did not and the binomial distribution table is:

Binomial distribution (n=10, p=0.303)

 f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0271 0.0271

1 0.1176 0.1447

2 0.2301 0.3748

3 0.2668 0.6416

4 <u>0.2029</u> 0.8445

5 0.1059 0.9504

6 0.0384 0.9888

7 0.0095 0.9983

8 0.0016 0.9998

9 0.0002 1.0000

10 0.0000 1.0000

<u>The probability is 0.2029 or 20.29%</u>

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x Pr[X = x] Pr[X ≤ x]

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1 0.0294 0.0319

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