The 2.88 minutes have to reduce the standard deviation to increase the efficiency.
According to the statement
we have given that the duration of the trip is 2 hours and 15 minutes. And we have to how much standard deviation be reduced to increase the operational efficiency.
So, For this purpose, we know that the
Let X be the time taken to reach the resort. Then it is given that the
N (μ , σ).
Here μ is 2 hours then
μ = 2*60 = 120 minutes.
And
Let σ be the reduced standard deviation such that new distributions becomes N (μ , σ)
Then P ( x <135)
here 135 is the minute format of 2 hours and 15 min.
So,
P ( x <135) = 0.95
And
P [ (x -120 / σ) < (135 -120 / σ)] = 0.95
P [ Z < ( 15 / σ)] = 0.95
And after solving it the
σ = 9.22 min.
i.e. 12 - 9.22 = 2.88 min.
So, The 2.88 minutes have to reduce the standard deviation to increase the efficiency.
Learn more about Standard deviation here
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Question:
The time it takes for a ferry to reach a summer resort from the mainland is Normally distributed with mean 2 hours and standard deviation 12 minutes. The advertised duration of the trip is 2 hours and 15 minutes. The company operating the ferry would like to increase its operational efficiency by reducing the standard deviation so that they are on time 95% of the time. How much do they have to reduce the standard deviation
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