Suppose that the weights of airline passenger bags are normally distributed with a mean of 48.08 pounds and a standard deviation of 3.13 pounds.
A)Let X represent the weight of a randomly selected bag. For what value of c is P(E(X) - c < X < E(X) + c)=0.84? Give your answer to four decimal places.
B)Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 16 that weigh less than 50 lbs? Give your answer to four decimal places.
C)Assuming the weights of individual bags are independent, what is the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16? Give your answer to four decimal places.
A) For this question, first, we know that E(X) = 48.08 pounds because this is the mean of the distribution. We want to find the value of c such that 84% of the data falls within the interval (48.08 - c, 48.08 + c). We can use the Z-tables to find this value of c by converting the problem to a standard normal distribution. To do this, we have:
P(Z < (X−48.08)/3.13) = 0.84/2 + 0.5 = 0.92
We can look up 0.92 in a Z-table of standard normal distribution, and we get a Z score of 1.4051. Then we can set up the equation:
1.4051 = (X - 48.08) / 3.13
Solve for X:
X = 1.4051 * 3.13 + 48.08 = 52.429313
Now we have c:
c = 52.429313 - 48.08 = 4.349313
So, the value of c is 4.3493.
B) To answer this question, we must first find the probability that a single bag weighs less than 50 pounds. We can do this using the Z-score and the standard normal distribution. To find this probability, we can first find the Z-score:
Z = (50 - 48.08) / 3.13 = 0.614057
Using the Z-table, we can find the probability corresponding to this Z-score:
P(X < 50) = P(Z < 0.614057) ≈ 0.7304
Now we are looking for the expected number of bags out of a sample of 16 that weigh less than 50 lbs. This is simply the probability that a single bag weighs less than 50 lbs multiplied by the total number of bags in the sample:
Expected number of bags = 0.7304 * 16 = 11.6864
So, the expected number of bags out of 16 that weigh less than 50 lbs is 11.6864 bags.
C) To answer this question, we will use the binomial distribution. The binomial distribution is used when there are two possible outcomes for each event (in this case, each bag weighs less than 50 lbs or it doesn't), and we want to find the probability of a certain number of successes (bags weighing less than 50 lbs) in a fixed number of trials (the sample of 16 bags). In this case, the binomial distribution formula is:
P(X ≤ k) = Σ (nCk * p^k * (1-p)^(n-k))
where n = 16 (the sample size), k = 12 (the number of successes), p = 0.7304 (the probability of a single success), and nCk is the number of combinations of n things taken k at a time.
We will compute this probability using the cumulative approach:
P(X ≤ 12) = P(X=0) + P(X=1) + ... + P(X=12)
Using the binomial distribution formula:
P(X ≤ 12) ≈ 0.0140
So, the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16 is approximately 0.0140, or 1.40%.
A) To find the value of c, we need to calculate the z-score corresponding to the 0.84 probability from the standard normal distribution.
Using the formula z = (X - μ) / σ, where X is the random variable, μ is the mean, and σ is the standard deviation, we can rearrange the equation to solve for X:
P(E(X) - c < X < E(X) + c) = 0.84
E(X) - c = μ - c = 48.08 - c
E(X) + c = μ + c = 48.08 + c
Now we have:
P(48.08 - c < X < 48.08 + c) = 0.84
Converting the bounds of inequality into z-scores:
P((48.08 - c - μ) / σ < Z < (48.08 + c - μ) / σ) = 0.84
Looking up the z-score with a cumulative probability of 0.84 (to find the corresponding Z in the standard normal distribution table), we find z = 0.9945.
Substituting the values into the equation:
(48.08 + c - 48.08) / 3.13 = 0.9945
Simplifying:
c / 3.13 = 0.9945
Solving for c:
c = 0.9945 * 3.13
c = 3.1095 (rounded to four decimal places)
Therefore, the value of c is approximately 3.1095 pounds.
B) Since the weights of individual bags are independent, the proportion of bags weighing less than 50 lbs follows a binomial distribution, with n = 16 and p = P(X < 50). We can calculate the expected value using the formula:
E(X) = n * p
First, we need to calculate the probability P(X < 50) by converting the weight into a z-score:
z = (X - μ) / σ
z = (50 - 48.08) / 3.13
Using the z-score table, we find that the cumulative probability for a z-score of 0.6127.
Therefore, the probability of a bag weighing less than 50 lbs is P(X < 50) = 0.6127.
Now, we can calculate the expected value:
E(X) = 16 * 0.6127
E(X) = 9.8032 (rounded to four decimal places)
Therefore, the expected number of bags out of a sample of 16 that weigh less than 50 lbs is approximately 9.8032.
C) To calculate the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16, we need to calculate the cumulative probability of P(X ≤ 12) from the binomial distribution with n = 16 and p = 0.6127.
Using statistical software or a calculator, we can find that P(X ≤ 12) ≈ 0.2376.
Therefore, the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16 is approximately 0.2376.
A) To find the value of c for which P(E(X) - c < X < E(X) + c) = 0.84, we need to calculate the confidence interval around the mean.
The confidence interval formula for a normal distribution is given by:
P(E(X) - c < X < E(X) + c) = 2 * P(Z < c/sigma) - 1
Where Z is the standard normal variable and sigma is the standard deviation.
Since we want the probability to be 0.84, we can set up the equation:
2 * P(Z < c/3.13) - 1 = 0.84
We can solve this equation by looking up the value of c/3.13 in the standard normal distribution table. Going through the table, we find that P(Z < 1.0057) is approximately 0.8415.
So, c/3.13 = 1.0057.
Solving for c, we multiply both sides by 3.13:
c = 3.13 * 1.0057
Therefore, c ≈ 3.1701 (rounded to four decimal places).
B) To find the expected number of bags out of a sample of 16 that weigh less than 50 lbs, we need to calculate the probability of an individual bag weighing less than 50 lbs and multiply it by the sample size.
The probability that an individual bag weighs less than 50 lbs can be found using the cumulative distribution function (CDF) of the normal distribution:
P(X < 50) = P(Z < (50 - 48.08) / 3.13)
= P(Z < 0.6140)
We can look up the value in the standard normal distribution table and find that P(Z < 0.6140) is approximately 0.7291.
So, the probability of an individual bag weighing less than 50 lbs is 0.7291.
Now, we can calculate the expected number of bags using the formula:
Expected number of bags = Probability of an individual bag < 50 lbs * Sample size
Expected number of bags = 0.7291 * 16
Therefore, the expected number of bags out of a sample of 16 that weigh less than 50 lbs is approximately 11.6656 (rounded to four decimal places).
C) To find the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16, we need to calculate the cumulative probability.
The formula for the cumulative probability is:
P(X ≤ 12) = P(X = 0) + P(X = 1) + ... + P(X = 12)
We can use the binomial probability formula to calculate each individual probability:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success (probability of an individual bag weighing less than 50 lbs), and n is the sample size.
Using this formula, we can compute each individual probability from X = 0 to X = 12 and sum them up:
P(X ≤ 12) = P(X = 0) + P(X = 1) + ... + P(X = 12)
= ∑[k=0 to 12] (C(16, k) * 0.7291^k * (1 - 0.7291)^(16 - k))
This calculation involves adding up thirteen terms, which can be quite time-consuming. Therefore, it is recommended to use a statistical software or calculator to obtain the result accurately.
Using a software or calculator, we can find that the probability that 12 or fewer bags weigh less than 50 pounds in a sample of size 16 is approximately 0.8921 (rounded to four decimal places).