Answer:
0.6554 is the probability that the mean number of gallons of fuel needed to take off for a randomly selected sample of 40 jumbo jets will be less than 3950 gallons.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4000 gallons
Standard Deviation, σ = 125 gallons
Sample size, n = 40
We are given that the distribution of amount of fuel is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Central limit theorem:
As the sample size increases, the distribution of sample mean has a similar popular distribution shape.
P(sample of 40 jumbo jets will be less than 3950 gallons)
P(x < 3950)
[tex]P( x < 3950) = P( z < \displaystyle\frac{3950 - 4000}{125}) = P(z < -0.4)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 3950) = 0.6554 = 65.54\%[/tex]
0.6554 is the probability that the mean number of gallons of fuel needed to take off for a randomly selected sample of 40 jumbo jets will be less than 3950 gallons.
The probability that the number of gallons of fuel needed to take off for a randomly selected sample of 40 jumbo jets will be less than 3950 gallons is 0.66.
it is given that
Mean μ= 4000 gallons
Standard deviation σ = 125 gallons
Number of trials x = 3950 gallons
Z-score = (x-μ)/σ
Z-score = (3950-4000)/125
Z-score = -0.4
So probbaility P(x<3950) = P(z<-0.4)
From the standard normal table,
P(x<3950) = 0.66
Therefore, the probability that the number of gallons of fuel needed to take off for a randomly selected sample of 40 jumbo jets will be less than 3950 gallons is 0.66.
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