Respuesta :
Answer:
There is a 33.72% probability that the total weight of the passengers exceeds 4500 pounds.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
There are 22 passengers. Passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. This means that [tex]\mu = 22*190 = 4180, \sigma = 22*35 = 770[/tex].
What is the approximate probability that the total weight of the passengers exceeds 4500 pounds?
This probability is 1 subtracted by the pvalue of Z when [tex]X = 4500[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4500 - 4180}{770}[/tex]
[tex]Z = 0.42[/tex]
[tex]Z = 0.42[/tex] has a pvalue of 0.6628.
This means that there is a 1-0.6628 = 0.3372 = 33.72% probability that the total weight of the passengers exceeds 4500 pounds.
The approximate probability that the total weight of the passengers exceeds 4500 pounds is 0.3372 = 33.72%
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For this case, we can take:
X = weight of a passenger in the considered flights
Then, by the specified data, we get:
Average weight = [tex]\mu = 190 \: \rm pounds[/tex]
Standard deviation of weight = [tex]\sigma = 35 \: \rm pounds[/tex]
And, [tex]X \sim N(\mu = 190, \sigma = 35)[/tex]
For the commuter plane that carries 22 passengers, let we have:
Y = total weight of all 22 passengers = [tex]22X[/tex]
Then, we get:
[tex]\mu_Y = E(Y) = E(22X) = 22E(X) = 22\times 190 = 4180[/tex]
and
[tex]\sigma_Y = \sqrt{Var(Y)} = \sqrt{Var(22X)} = \sqrt{22^2Var(X)} = 22\sqrt{35^2} = 770[/tex]
(it is because [tex]E(aX) = aE(X) , Var(aX) = a^2Var(X)[/tex] )
Since Y is scaled version of X, thus, it is also normally distributed, or:
[tex]Y \sim N(\mu_Y = 4180, \sigma_Y = 770)[/tex]
Then, the probability that value of Y exceeds 4500 pounds can be symbolically written as:
[tex]P(Y > 4500)[/tex]
It can be rewritten as:
[tex]P(Y > 4500) = 1- P (Y \leq 4500)[/tex]
Using the standard normal distribution, we get the value of this probability as:
[tex]P(Y > 4500) = 1- P (Y \leq 4500) = 1 - P(Z \leq z = \dfrac{y-\mu_Y}{\sigma_Y})\\\\P(Y > 4500) = 1 - P(Z \leq \dfrac{4500-4180}{770}) \approx 1-P(Z \leq 0.42)[/tex]
From z-tables, the p-value for Z = 0.42 is obtained as 0.6628
Thus, we get:
[tex]P(Y > 4500) \approx 1-P(Z \leq 0.42) = 1 - 0.6628 = 0.3372[/tex]
Thus, the approximate probability that the total weight of the passengers exceeds 4500 pounds is 0.3372 or 33.72%
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
