Question

Suppose that the weights of airline passenger bags are normally distributed with a mean of 49.02 pounds and a standard deviation of 3.83 pounds.
a) What is the probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds? Give your answer to four decimal places.

b) Let X represent the weight of a randomly selected bag. For what value of c is P(E(X) - c < X < E(X) + c)=0.82? Give your answer to four decimal places. Incorrect: Your answer is incorrect.

c) Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 17 that weigh less than 50 lbs? Give your answer to four decimal places.

d) Assuming the weights of individual bags are independent, what is the probability that 11 or fewer bags weigh less than 50 pounds in a sample of size 17? Give your answer to four decimal places.

Answer 1

Answer:

a) P(X<50)=  0.60257

b) IcI= 54.1522

c) E(X)= 10.2 bags

d) P(X≤11) = 0.7361

Step-by-step explanation:

Hello!

The study variable is X: the weight of an airline passenger bag.

X~N(μ; δ²)

μ= 49.02

δ= 3.83

a) What is the probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds?

P(X<50)= P(Z<$\frac{50 - 49.02}{3.83}$)=

P(Z<0.2558)≅ P(Z<0.26)= 0.60257

b) Let X represent the weight of a randomly selected bag.

For what value of c is P(E(X)-c < X < E(X) + c)

The normal distribution is symmetric and centered in the mean E(X)= μ and IcI represents a number, you can say that between -c and +c there is 0.82 of probability and the rest 1 - 0.82= 0.18 is divided between the two tails left out of the interval E(X)-c < X < E(X) + c.

If 1 - α= 0.82, then α= 0.18 and α/2= 0.09.

The cumulative probability to -c is α/2= 0.09 and the cumulative probability to +c is (α/2) + (1 - α)= 0.09 + 0.82= 0.91

(see graphic)

Now that you know what is the cumulative probability for each tail, you can calculate the value of IcI, either tail is the same, the module value won't change:

P(X < c)= 0.91

There is no need to write E(X) since the mean is the center of the distribution and we already took that into account when deducing the cumulative probabilities.

Standardize it:

P(Z < d)= 0.91 ⇒ Look for 0.91 in the body of the standard normal table to find the value that corresponds to a probability of 0.91

Where:

d=  (c - μ)

δ

d= 1.34

Now you have to reverse the standardization

1.34=  (c - 49.02)

3.83

c - 49.02= 1.34 * 3.83

c= (1.34*3.83)+49.02

c= 54.1522

c) Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 17 that weigh less than 50 lbs? Give your answer to four decimal places.

Now the study variable has changed, we are no longer interested in the weight of the bag but in the number of bags that meet a certain characteristic (weigh less than 50 pounds). The new study variable is:

X: Number of bags that weigh less than 50 pounds in a sample of 17 bags.

The bags are independent, the number of trials of the experiment is fixed n=17, there are only two possible outcomes success "the bag weighs less than 50 pounds" and failure "the bag weighs at least 50 pounds" and the probability of success is the same trough all the trial, so we can say that the new variable has a binomial distribution, symbolically:

X~Bi(n;p)

Since we are basing this new variable in the same population of bags, we already calculated the probability of success of the experiment in par a) P(X<50)= p= 0.60257≅ 0.60

For a variable with a binomial distribution, the expected value is E(X)=np

E(X)=17*0.60= 10.2 bags

We expect that 10.2 bags weigh less than 50 pounds.

d) Assuming the weights of individual bags are independent, what is the probability that 11 or fewer bags weigh less than 50 pounds in a sample of size 17? Give your answer to four decimal places.

For this item we will be working with the same variable as in c)

n= 17 p=0.60

P(X≤11) = 0.7361

I hope it helps!