I'm having trouble with this word problem:
Priority Mail is the U.S. Postal Service's alternative to commercial express mail companies such as FedEx. An article in The Wall Street Journal presented some interesting conclusions comparing Priority Mail shipments with the much less expensive first-class shipments. When comparing shipments intended for delivery in three days, first-class deliveries failed to deliver on time 19% of the time, while Priority Mail failed 33% of the time. Note that at the time of the article, first-class deliveries started as low as $0.34 and Priority Mail started at $3.50.
If 10 items are to be shipped first-class to 10 different destinations claimed to be in a three-day delivery location, what is the probability that
a. 0 items will take more than three days?
b. exactly 1 will take more than three days?
c. 2 or more will take more than three days?
d. What are the mean and the standard deviation of the probability distribution?
Do you know where to find this answer?
To solve this word problem, we need to calculate the probabilities based on the given information. Let's start by understanding the required probabilities step by step.
a. Probability that 0 items will take more than three days:
Since first-class deliveries fail to deliver on time 19% of the time, the probability that one item will take more than three days is 0.19. Therefore, the probability that 0 items will take more than three days is the complement of one item taking more than three days, which is 1 - 0.19 = 0.81.
b. Probability that exactly 1 item will take more than three days:
To calculate this probability, we need to consider the binomial probability formula. In this case, we have 10 trials (10 items shipped), and the probability of success (an item taking more than three days) is 0.19. The probability of exactly one success can be calculated using the formula:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X = k) represents the probability of exactly k successes
- nCk represents the binomial coefficient, which is calculated as n! / (k! * (n-k)!)
- p represents the probability of success in a single trial
- (1-p) represents the probability of failure in a single trial
- n represents the total number of trials
For this problem, we have n = 10, k = 1, and p = 0.19. Plugging these values into the formula, we get:
P(X = 1) = (10C1) * 0.19^1 * (1-0.19)^(10-1)
You can use a calculator or a binomial probability calculator to find the answer. The result is approximately 0.384.
c. Probability that 2 or more items will take more than three days:
To calculate this probability, we need to consider the complement of the probabilities calculated in part (a) and part (b).
P(2 or more taking more than three days) = 1 - P(0 taking more than three days) - P(1 taking more than three days)
Substituting the calculated probabilities, we get:
P(2 or more taking more than three days) = 1 - 0.81 - 0.384
Calculating this, we find that the probability is approximately 0.194.
d. Mean and standard deviation of the probability distribution:
To calculate the mean and standard deviation, we need to use the properties of a binomial distribution.
For a binomial distribution, the mean (μ) can be calculated using the formula:
μ = n * p
Where n represents the number of trials and p represents the probability of success in a single trial. In this problem, n = 10 and p = 0.19, so the mean is:
μ = 10 * 0.19 = 1.9
To calculate the standard deviation (σ), we use the formula:
σ = sqrt(n * p * (1-p))
Plugging in the values, we get:
σ = sqrt(10 * 0.19 * (1 - 0.19))
Calculating this, we find that the standard deviation is approximately 1.246.
So, the mean of the probability distribution is 1.9, and the standard deviation is 1.246.