Respuesta :
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 = [tex] \frac{s}{ \sqrt{n} }= \frac{5.3}{ \sqrt{100} }=0.53 [/tex]
Now we have to convert the weight to z-score
[tex]z= \frac{x-u}{ \frac{s}{ \sqrt{n} } } [/tex]
[tex]z= \frac{21-19.4}{0.53}=3.018 [/tex]
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%
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 = [tex] \frac{s}{ \sqrt{n} }= \frac{5.3}{ \sqrt{100} }=0.53 [/tex]
Now we have to convert the weight to z-score
[tex]z= \frac{x-u}{ \frac{s}{ \sqrt{n} } } [/tex]
[tex]z= \frac{21-19.4}{0.53}=3.018 [/tex]
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:
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 = [tex]\frac{SD}{\sqrt{n} } =0.53[/tex]
To find the probability that X>2100/100 lbs
=P(X<21[tex]\frac{21-19.4}{0.53} =3.019[/tex]) =P(Z<3.019)
=0.5+0.4987
0.9987
