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Nimfa-mama [501]
3 years ago
12

Number of passengers who arrive at the platform in an amtrack train station for the 2 pm train on a saturday is a random variabl

e with a mean of 180 and a standard deviation of 25.
a.with what probability can we assert that there will be between 80 and 280 passengers?
b.with what probability can we assert that there will be between 130 and 230 passengers?
Mathematics
1 answer:
alexgriva [62]3 years ago
8 0
Assuming a normal distribution  we find the standardized z scores for  a:-

z1 = (80 - 180)  / 25  =  = -100/25  = -4
z2 =  (280-180) / 25  =  4
Required P( -4 < z < 4)  from the tables is  >99.9%

b
z1 = 130-180 / 25 =  -2
z2 =  230-180 / 25  = 2

from tables probability is 2* 0.4773  =  95.46 %
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Answer:

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Step-by-step explanation:

We have that the speed is the distance divided by the time. Mathematically, that is

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(a) how much time is spent on the trip and

The peron's average speed is 77.8 km/h, which means that s = 77.8

The person distance traveled is:

22 min is 22/60 = 0.37h.

So for  the time t1, the person traveled at a speed of 89.5 km/h. Which has a distance of 89.5*t1.

For 0.37h, the person was at a stop, so she did not travel. This means that the total distance is

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The total time is the time traveling t and the stoppage time 0.37. So

t = t1 + 0.37

We want to find t1, which is the time that the person was driving.

So

s = \frac{d}{t}

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77.8t1 + 77.8*0.37 = 89.5t1

11.7t1 = 28.786

t1 = \frac{28.786}{11.7}

t1 = 2.46

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The person traveled for 2.83 hours.

(b) how far does the person travel?

The person traveled 2.46 hours at an average speed of 77.8 km/h. So

s = \frac{d}{t}

77.8 = \frac{d}{2.83}

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