Today, the waves are crashing onto the beach every 5.3 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.3 seconds. Round to 4 decimal places where possible.The probability that the wave will crash onto the beach between 0.4 and 4.7 seconds after the person arrives is P(0.4 < x < 4.7) = The probability that it will take longer than 2.96 seconds for the wave to crash onto the beach after the person arrives is P(x ≥≥ 2.96) = Find the maximum for the lower quartile.  seconds.

Question
Asked Nov 2, 2019

Today, the waves are crashing onto the beach every 5.3 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.3 seconds. Round to 4 decimal places where possible.

The probability that the wave will crash onto the beach between 0.4 and 4.7 seconds after the person arrives is P(0.4 < x < 4.7) = 

The probability that it will take longer than 2.96 seconds for the wave to crash onto the beach after the person arrives is P(x ≥≥ 2.96) = 

Find the maximum for the lower quartile.  seconds.

check_circleExpert Solution
Step 1

Let X be the random variable uniformly distributed between 0 and 5.3 seconds.

Denote X as the time from a person arrives at the shoreline until a crashing wave is observed.

Thus, the probability density function of X is:

1
Г(х -х)-
b-a
1
5.3 0
1
5.3
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1 Г(х -х)- b-a 1 5.3 0 1 5.3

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Step 2

The probability that the wave will crash on to the beach between 0.4 and 4.7 seconds after the person arrives is calculated as follows:

4.7
P(0.4 X 4.7)
-dx
5.3
0.4
1
0.4
5.3
1
(4.7-0.4)
5.3
4.3
5.3
0.81132
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Image Transcriptionclose

4.7 P(0.4 X 4.7) -dx 5.3 0.4 1 0.4 5.3 1 (4.7-0.4) 5.3 4.3 5.3 0.81132

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Step 3

The probability that it will take longer than 2.96 seconds for the wave to cr...

53
1
P(X>2.96)=
5.3
2.96
1
296
5.3
1
(5.3-2.96)
5.3
2.34
5.3
0.44151
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53 1 P(X>2.96)= 5.3 2.96 1 296 5.3 1 (5.3-2.96) 5.3 2.34 5.3 0.44151

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