Suppose that 10% of the fields in a given agricultural area with about 2,500 fields are infested with the sweet potato white fly
1 answer:
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Answer:
a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes
c) 0.2971 = 29.71% 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:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 8.9 minutes and a standard deviation of 2.9 minutes.
This means that
Sample of 37:
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?
320/37 = 8.64865
Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So
By the Central Limit Theorem
has a p-value of 0.2981
0.2981 = 29.81% 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?
275/37 = 7.4324
Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So
has a p-value of 0.001
1 - 0.001 = 0.999
0.999 = 99.9% 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?
Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So
0.2981 - 0.0010 = 0.2971
0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes