The arnual per capita consumption of bottied water was 34.4 gallons. Assume that the per copita consumption of bottled water is approvimately norrmaly distributhed with a mein of 34.4 and a standard teviation of 13 gallons. a. What is the probablity that someone consumed more than 39 gallons of bottlod water? b. What is the probability that someone consumed between 20 and 30 gallons of botted water? c. What is the probability that someone consumed less than 20 gallons of botilied waster? d. 99% of people consumed less than how many gallons of botted water? a. The probability that someone consumed more than 39 gallians of bottied water is (Round to four decimal places as needed.)

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To find the probability that someone consumed more than 39 gallons of bottled water, we can use the normal distribution formula.

Let X be the per capita consumption of bottled water. We can calculate the z-score for 39 gallons using the formula:

z = (X - μ) / σ
z = (39 - 34.4) / 13

Using a standard normal distribution table or calculator, we can find the probability corresponding to this z-score:

P(X > 39) = P(Z > (39 - 34.4) / 13)
P(X > 39) = P(Z > 0.3538)

The probability that someone consumed more than 39 gallons of bottled water is approximately 1 - 0.6368 = 0.3632.

a. The probability that someone consumed more than 39 gallons of bottled water is approximately 0.3632.

b. To find the probability that someone consumed between 20 and 30 gallons of bottled water, we need to calculate the z-scores for both values and find the area between them.

z_20 = (20 - 34.4) / 13
z_30 = (30 - 34.4) / 13

Using the standard normal distribution table or calculator, we can find the probabilities corresponding to these z-scores:

P(20 < X < 30) = P(z_20 < Z < z_30)
P(20 < X < 30) = P((-1.1077 < Z < -0.3385)

The probability that someone consumed between 20 and 30 gallons of bottled water is approximately 0.4084 - 0.3694 = 0.039.

b. The probability that someone consumed between 20 and 30 gallons of bottled water is approximately 0.039.

c. To find the probability that someone consumed less than 20 gallons of bottled water, we can calculate the z-score for 20 gallons:

z = (20 - 34.4) / 13
z = -1.1077

Using the standard normal distribution table or calculator, we can find the probability corresponding to this z-score:

P(X < 20) = P(Z < -1.1077)

The probability that someone consumed less than 20 gallons of bottled water is approximately 1 - 0.3694 = 0.6306.

c. The probability that someone consumed less than 20 gallons of bottled water is approximately 0.6306.

d. To find the value that corresponds to the 99th percentile, we need to find the z-score that corresponds to this percentile and then convert it back to the original scale using the mean and standard deviation.

The z-score that corresponds to the 99th percentile is approximately 2.3263. Rearranging the formula for the z-score:
z = (X - μ) / σ

We can solve for X:

X = z * σ + μ
X = 2.3263 * 13 + 34.4

X ≈ 63.45

99% of people consumed less than approximately 63.45 gallons of bottled water.

d. 99% of people consumed less than approximately 63.45 gallons of bottled water.