So ummm anyone down to please help me I need to turn in the assignment tomorrow
1 answer:
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Answer:
y = (5/4)2^x
Step-by-step explanation:
The function value increases by a factor of 40/10 = 4 when x increases by 2. The function can be written as ...
y = (reference value)·(growth factor)^((x -reference)/(change in x for growth factor))
y = 10·4^((x-3)/2) . . . . . . using point (3, 10) as a reference
This can be simplified to ...
y = 10·2^(x -3) = 10/8·2^x
y = (5/4)2^x
Answer:
a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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:
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that .
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
We are working with a sample mean of 37 jets. So we have that:
Total time of 320 minutes for 37 jets, so
This probability is the pvalue of Z when . So
has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?
Total time of 275 minutes for 37 jets, so
This probability is subtracted by the pvalue of Z when
has a pvalue of 0.0548.
There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?
Total time of 320 minutes for 37 jets, so
Total time of 275 minutes for 37 jets, so
This probability is the pvalue of Z when subtracted by the pvalue of Z when
.
So:
From a), we have that for , we have
, that has a pvalue of 0.7422.
From b), we have that for , we have
, that has a pvalue of 0.0548.
So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.