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Molodets [167]
3 years ago
13

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.4 minutes and a standard deviation of 3.5

minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway.What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
Mathematics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

0.6672 is the required probability.            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.4 minutes

Standard Deviation, σ = 3.5 minutes

We are given that the distribution of distribution of taxi and takeoff times is a bell shaped distribution that is a normal distribution.

According to central limit theorem the sum measurement of n is normal with mean \mu and standard deviation \sigma\sqrt{n}

Sample size, n = 37

Standard Deviation =

=\sigma\times \sqrt{n} = 3.5\times \sqrt{37}=21.28

P(taxi and takeoff time will be less than 320 minutes)

P( \sum x < 320) = P( z < \displaystyle\frac{320 - 37(8.4)}{21.28}) = P(z < 0.4323)

Calculation the value from standard normal z table, we have,  

P(\sum x < 320) =0.6672 = 66.72\%

0.6672 is the probability  that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

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Find (f −1)'(a).<br><br> f(x) = 4 + x2 + tan(πx/2), −1 &lt; x &lt; 1, a = 4
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Answer:

The  solution  is   (f^{-1})'(a) = \frac{2}{\pi }

Step-by-step explanation:

From the question we are told that

      The  function is  f(x) =  4 +  x^2  + tan [\frac{ \pi x}{2} ] ,     -1 <  x  <  1  a =  4

Here we are told find  (f^{-1}) (a)

Let equate

     f(x) =  a

So  

      4 +  x^2  + tan[\frac{\pi x }{2} ] =  4

       x^2  + tan[\frac{\pi x }{2} ]  =  0

For the equation above to be valid x must be equal to 0

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Differentiating  f(x)

     f(x)'  =  0 + 2x  + sec^2 (\frac{\pi x}{2} )\cdot \frac{\pi}{2}

Now

    since 0 =  f^{-1} (4)

We have

      f(0)'  =  0 +  \frac{\pi }{2}  sec^2 (0)

       f(0)'  = \frac{\pi }{2}

Now  

        (f^{-1})'(a) =  \frac{1}{(\frac{\pi}{2} )}

        (f^{-1})'(a) = \frac{2}{\pi }

       

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3 years ago
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