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777dan777 [17]
1 year ago
11

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

10 second period. Compute the probability that more than 3 messages will arrive during a 30-second interval.
Mathematics
1 answer:
NARA [144]1 year ago
7 0

The probability that more than 3 messages will arrive during a 30-second interval is P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}.

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\

And then substitute the values in it then

Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}.

Learn more about probability here

brainly.com/question/24756209

#SPJ4

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Step One
Since (1) and (2) both have y isolated on their respective right sides, they can be equated.

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The box-and-whisker plot below shows the numbers of text messages received in one day by students in the seventh and eighth grad
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<h3><u>Answer with explanation:</u></h3>

From the box and whisker plot of seventh grade we have:

The minimum value =6

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and maximum value = 26

From the box and whisker plot of eighth grade we have:

The minimum value =22

First quartile or lower quartile i.e. Q_1 = 26

Median or second quartile i.e. Q_2 = 30

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a)

The overlap of the two sets of data is as follows.

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IQR is calculated as the difference of the Upper quartile and the lower quartile

i.e. Q_3-Q_1

so, IQR of seventh grade is:

22-14=8

IQR of seventh grade=8

IQR of eighth grade is:

34-26=8

Hence, IQR of eighth grade=8

c)

The difference of the median of the two data sets is:

30-18=12

Hence, the difference of median is: 12

d)

As the IQR of both the sets is same i.e. 8.

Hence, the number that must be multiplied by IQR so that it is equal to the difference between the medians of the two sets is:

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