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

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 = $\frac{s}{ \sqrt{n} }= \frac{5.3}{ \sqrt{100} }=0.53$

Now we have to convert the weight to z-score

$z= \frac{x-u}{ \frac{s}{ \sqrt{n} } }$
 
$z= \frac{21-19.4}{0.53}=3.018$

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%

Answer 2

Answer:

0.9987

Step-by-step 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 = $\frac{SD}{\sqrt{n} } =0.53$

To find the probability that X>2100/100 lbs

=P(X<21$\frac{21-19.4}{0.53} =3.019$) =P(Z<3.019)

=0.5+0.4987

0.9987