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.)
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.
To solve this problem, we will use the z-score formula to standardize the values and then look up the probabilities in the standard normal distribution table.
a. To find the probability that someone consumed more than 39 gallons of bottled water, we need to calculate the z-score and then find the corresponding probability.
The formula to find the z-score is:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
Calculating the z-score for x = 39 gallons:
z = (39 - 34.4) / 13
z = 0.3538
Using the standard normal distribution table, we can find the probability associated with a z-score of 0.3538. The table provides the area to the left of the z-score, so to find the probability of consuming more than 39 gallons, we need to subtract the area to the left of the z-score from 1.
P(X > 39) = 1 - P(X ≤ 39)
= 1 - 0.6368
= 0.3632
Therefore, the probability that someone consumed more than 39 gallons of bottled water is 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 each value and then find the difference in their probabilities.
Calculating the z-score for x = 20 gallons:
z1 = (20 - 34.4) / 13
z1 = -1.1077
Calculating the z-score for x = 30 gallons:
z2 = (30 - 34.4) / 13
z2 = -0.3385
Using the standard normal distribution table, we can find the probabilities associated with z1 and z2:
P(X ≤ 20) = 0.1368
P(X ≤ 30) = 0.3694
To find the probability of consuming between 20 and 30 gallons, we subtract the probabilities:
P(20 ≤ X ≤ 30) = P(X ≤ 30) - P(X ≤ 20)
= 0.3694 - 0.1368
= 0.2326
Therefore, the probability that someone consumed between 20 and 30 gallons of bottled water is 0.2326.
c. To find the probability that someone consumed less than 20 gallons of bottled water, we need to calculate the z-score for x = 20 gallons and find the corresponding probability.
Calculating the z-score for x = 20 gallons:
z = (20 - 34.4) / 13
z = -1.1077
Using the standard normal distribution table, we can find the probability associated with a z-score of -1.1077, which represents consuming less than 20 gallons:
P(X < 20) = 0.1368
Therefore, the probability that someone consumed less than 20 gallons of bottled water is 0.1368.
d. To find the value at which 99% of people consumed less than, we need to find the z-score that corresponds to a cumulative probability of 0.99.
Using the standard normal distribution table, we can find the z-score that corresponds to a cumulative probability of 0.99:
z = 2.3263
Now, we can use the z-score formula to find the value:
z = (x - μ) / σ
Plugging in the known values:
2.3263 = (x - 34.4) / 13
Solving for x:
2.3263 * 13 = x - 34.4
30.2409 = x - 34.4
x = 64.6409
Therefore, 99% of people consumed less than 64.6409 gallons of bottled water.