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astra-53 [7]
3 years ago
5

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 00 and 66

minutes. find the probability that a randomly selected passenger has a waiting time greater than greater than 2.252.25 minutes.
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

You refer to a rectangular distribution. 

I.e. Area ∝ probability 

so, length x height = area 

=> 6 x height = 1 

so, height = 1/6 

Therefore, greater than 2.25 minutes => 8-2.25 = 5.75 minutes 

=> 5.75 x 1/6 = 0.9583

 

Since this is uniformly distributed, there is no need to find the height or standard deviation.

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Find the distance speed=72km/h time=45mins
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The distance is 54 km if the speed and time are 72 km/h and 45 minutes respectively.

<h3>What is distance?</h3>

Distance is a numerical representation of the distance between two items or locations. Distance refers to a physical length or an approximation based on other physics or common usage considerations.

We have:

Speed = 72 km/h

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We know,

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2 years ago
The following data are the distances between a sample of 20 retail stores and a large distribution center. The distances are in
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Variance = 1,227.27

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Step-by-step explanation:

To calculate these, we use the following formulas:

Mean = (sum of the values) / n

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Where;

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x = each value

Therefore, we have:

Sum of the values = 29 + 32 + 36 + 40 + 58 + 67 + 68 + 69 + 76 + 86 + 87 + 95 + 96 + 96 + 99 + 106 + 112 + 127 + 145 + 150 = 1,674

Mean = 1,674 / 20 = 83.70

Variance = ((29-83.70)^2 + (32-83.70)^2 + (36-83.70)^2 + (40-83.70)^2 + (58-83.70)^2 + (67-83.70)^2 + (68-83.70)^2 + (69-83.70)^2 + (76-83.70)^2 + (86-83.70)^2 + (87-83.70)^2 + (95-83.70)^2 + (96-83.70)^2 + (96-83.70)^2 + (99-83.70)^2 + (106-83.70)^2 + (112-83.70)^2 + (127-83.70)^2 + (145-83.70)^2 + (150-83.70)^2) / (20 - 1) = 23,318.20 / 19 = 1,227.27

Standard deviation = 1,227.27^0.5 = 35.03

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