Question

The distribution of the baggage weights for passengers using a particular airline has a mean of 19.4 lbs and a standard deviation of 5.3 lbs. what is the probability that for (a random sample of) 100 passengers … the total luggage weight is less than 2,100 lbs?

Answer 1

0.9987

Explanation:

Given that X, The distribution of the baggage weights for passengers using a particular airline has a mean of 19.4 lbs and a standard deviation of 5.3 lbs.

For a sample of size 100, we have

n=100

Hence std deviation of sample =
`(SD)/(√(n) ) =0.53`

To find the probability that X>2100/100 lbs

=P(X<21
`(21-19.4)/(0.53) =3.019`) =P(Z<3.019)

=0.5+0.4987

0.9987

Answer 2

We are to find the probability that the weight of total luggage for a sample of 100 passengers is less than 2100.

The mean weight of the luggage of passengers will be 2100/100 = 21.

So we have to find the probability of the mean weight to be less than 21.

Average weight = u = 19.4
Standard deviation = 5.3

Since we are dealing with a sample of 100. We will use the standard error.

Standard error =
`(s)/( √(n) )= (5.3)/( √(100) )=0.53`

Now we have to convert the weight to z-score


`z= (x-u)/( (s)/( √(n) ) )`


`z= (21-19.4)/(0.53)=3.018`

From z table we can find the probability of z being less than 3.018 is 99.87%.

Therefore, the probability that for (a random sample of) 100 passengers, the total luggage weight is less than 2,100 lbs is 99.87%