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.

Question

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.

Answer 1

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%.