A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag follows the continuous uniform distribution with low and high filling weights of 7.5 pounds and 11.6 pounds, respectively.
a. Calculate the expected value and the standard deviation of this distribution. (Do not round intermediate calculations. Round your "Expected value" to 2 decimal places and "Standard deviation" answer to 4 decimal places.)
b. Find the probability that the weight of a randomly selected bag is no more than 8.8 pounds. (Round your answer to 2 decimal places.)
c. Find the probability that the weight of a randomly selected bag is at least 8.3 pounds. (Round your answer to 2 decimal places.)
a. The expected value of a continuous uniform distribution can be calculated as the average of the low and high filling weights:
Expected value = (low + high) / 2 = (7.5 + 11.6) / 2 = 9.55 pounds
The standard deviation of a continuous uniform distribution can be calculated using the formula:
Standard deviation = (high - low) / sqrt(12)
Standard deviation = (11.6 - 7.5) / sqrt(12) ≈ 1.4285 pounds
b. To find the probability that the weight of a randomly selected bag is no more than 8.8 pounds, we need to calculate the area of the probability density function (PDF) below 8.8 pounds.
The probability density function of a continuous uniform distribution is a constant value between the low and high filling weights. Since the range of the distribution is 11.6 - 7.5 = 4.1 pounds, the PDF is 1/4.1 within this range.
To calculate the probability, we need to find the area under the PDF curve between 7.5 and 8.8 pounds:
Probability = (8.8 - 7.5) / 4.1 ≈ 0.3171
Therefore, the probability that the weight of a randomly selected bag is no more than 8.8 pounds is approximately 0.32.
c. To find the probability that the weight of a randomly selected bag is at least 8.3 pounds, we need to calculate the area of the probability density function (PDF) above 8.3 pounds.
The PDF is still 1/4.1 within the range of 7.5 and 11.6 pounds.
To calculate the probability, we need to find the area under the PDF curve above 8.3 pounds:
Probability = (11.6 - 8.3) / 4.1 ≈ 0.8049
Therefore, the probability that the weight of a randomly selected bag is at least 8.3 pounds is approximately 0.80.
a. To calculate the expected value (mean) and the standard deviation of a continuous uniform distribution, we can use the following formulas:
Expected value (mean) = (low + high) / 2
Standard deviation = (high - low) / sqrt(12)
Given that the low filling weight is 7.5 pounds and the high filling weight is 11.6 pounds, we can plug these values into the formulas:
Expected value = (7.5 + 11.6) / 2 = 19.1 / 2 = 9.55 pounds
Standard deviation = (11.6 - 7.5) / sqrt(12) = 4.1 / sqrt(12) ≈ 1.1864 pounds
Therefore, the expected value is 9.55 pounds and the standard deviation is approximately 1.1864 pounds.
b. To find the probability that the weight of a randomly selected bag is no more than 8.8 pounds, we need to calculate the cumulative distribution function (CDF) at 8.8 pounds.
The CDF of a continuous uniform distribution is given by:
CDF(x) = (x - low) / (high - low)
Plugging in the values, we have:
CDF(8.8) = (8.8 - 7.5) / (11.6 - 7.5) = 1.3 / 4.1 ≈ 0.3171
Therefore, the probability that the weight of a randomly selected bag is no more than 8.8 pounds is approximately 0.3171 or 31.71%.
c. To find the probability that the weight of a randomly selected bag is at least 8.3 pounds, we need to calculate the complementary cumulative distribution function (CCDF) at 8.3 pounds.
The CCDF of a continuous uniform distribution is given by:
CCDF(x) = 1 - CDF(x)
Plugging in the values, we have:
CCDF(8.3) = 1 - CDF(8.3) = 1 - ((8.3 - 7.5) / (11.6 - 7.5)) = 1 - 0.8 / 4.1 ≈ 0.8073
Therefore, the probability that the weight of a randomly selected bag is at least 8.3 pounds is approximately 0.8073 or 80.73%.
a. To calculate the expected value and standard deviation of the continuous uniform distribution, we can use the following formulas:
Expected Value (mean) = (low + high) / 2
Standard Deviation = (high - low) / sqrt(12)
Given that the low filling weight is 7.5 pounds and the high filling weight is 11.6 pounds, we can plug these values into the formulas to calculate:
Expected Value = (7.5 + 11.6) / 2 = 9.55 pounds (rounded to 2 decimal places)
Standard Deviation = (11.6 - 7.5) / sqrt(12) = 1.1897 pounds (rounded to 4 decimal places)
Therefore, the expected value is 9.55 pounds and the standard deviation is 1.1897 pounds.
b. To find the probability that the weight of a randomly selected bag is no more than 8.8 pounds, we need to calculate the z-score and then use a standard normal distribution table.
First, we calculate the z-score using the formula:
z = (x - mean) / standard deviation
where x is the given weight (8.8 pounds), mean is the expected value (9.55 pounds), and standard deviation is the standard deviation (1.1897 pounds).
z = (8.8 - 9.55) / 1.1897 ≈ -0.6300
Next, using the standard normal distribution table, we can find the probability corresponding to the z-score of -0.6300. This probability represents the area to the left of the z-score.
From the table, the probability corresponding to -0.63 is approximately 0.2632.
Therefore, the probability that the weight of a randomly selected bag is no more than 8.8 pounds is approximately 0.2632 (rounded to 2 decimal places).
c. To find the probability that the weight of a randomly selected bag is at least 8.3 pounds, we can use the complement rule.
The complement of "at least 8.3 pounds" is "less than 8.3 pounds". So, we need to find the probability that the weight is less than 8.3 pounds and subtract it from 1.
Using the same method as before, we find the z-score for 8.3 pounds:
z = (8.3 - 9.55) / 1.1897 ≈ -1.05
Looking up the z-score in the standard normal distribution table, we find the probability corresponding to -1.05 is approximately 0.1469.
Therefore, the probability that the weight of a randomly selected bag is at least 8.3 pounds is approximately 1 - 0.1469 = 0.8531 (rounded to 2 decimal places).