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dedylja [7]
3 years ago
14

a statistics professor finds that when he schedules an office hour at the 10:30a, time slot, an average of three students arrive

. Find the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive
Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
6 0

Answer:

The probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.

Step-by-step explanation:

Let <em>X</em> = number of students arriving at the 10:30 AM time slot.

The average number of students arriving at the 10:30 AM time slot is, <em>λ</em> = 3.

A random variable representing the occurrence of events in a fixed interval of time is known as Poisson random variables. For example, the number of customers visiting the bank in an hour or the number of typographical error is a book every 10 pages.

The random variable <em>X</em> is also a Poisson random variable because it represents the fixed number of students arriving at the 10:30 AM time slot.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.

The probability mass function of <em>X</em> is given by:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...,\ \lambda>0

Compute the probability of <em>X</em> = 2 as follows:

P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.0498\times 9}{2}=0.2241

Thus, the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.

Kaylis [27]3 years ago
3 0

Answer:

Probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.224.

Step-by-step explanation:

We are given that a statistics professor finds that when he schedules an office hour at the 10:30 am, time slot, an average of three students arrive.

Let X = number of students arriving at the 10:30 AM time slot.

<em>The above situation can be represented through Poisson distribution because a random variable representing the occurrence or arrival of events in a fixed period of time is known as Poisson random variables. </em>

The probability distribution function of Poisson distribution is given by;

P(X =x) = \frac{e^{-\lambda }\times \lambda^{x}  }{x!} ; x=0,1,2,3,.......

where, \lambda = arrival rate of students

As we know that the mean of Poisson distribution = \lambda

So, X ~ Poisson(\lambda=3)

Thus, the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is given by = P(X = 2)

       P(X = 2) = \frac{e^{-3 }\times 3^{2}  }{2!}

                     = \frac{e^{-3} \times 9 }{2} = 0.224

<em>Therefore, probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.224.</em>

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Now we use row operations to find the echelon form of the matrix:

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We substract to row 3, three times the row 1.

We substract to row 4, four times the row 1 and obtain the matrix

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2. We multiply the second row of the preview step by -1/4. We obtain the matrix

\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]

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We add to row 3, eight times the row 2.

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\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&-5&10&-155\end{array}\right]

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\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&0&0&-140\end{array}\right]

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