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3 years ago

Help me some one plz (only part b)

Mathematics

1 answer:

Slav-nsk [51]3 years ago

5 0

The school can expect to raise 1500p for charity.

<u>Explanation:</u>

(a)

The possibility space is solved in the image attached.

(b)

20p for each player

30p for lollipop

Number of people = 120

Charity amount = ?

Out of 12 outcomes, the probability of getting 7 is 3

So, probability of getting 7 for 120 people = $\frac{3}{12} X 120$

= 30

Money earned from the players = 20p X 120

= 2400p

Cost of 30 lollipops = 30p X 30

= 900p

Money received for charity = 2400p - 900p

= 1500p

Therefore, the school can expect to raise 1500p for charity.

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Birds arrive at a birdfeeder according to a Poisson process at a rate of six per hour.

m_a_m_a [10]

Answer:

a) time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b) $P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c) $P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

$P(T>t)= e^{-\lambda t}$

a. What is the expected time you would have to wait to see ten birds arrive?

The original rate for the Poisson process is given by the problem "rate of six per hour" and on this case since we want the expected waiting time for 10 birds we have this:

time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b. What is the probability that the elapsed time between the second and third birds exceeds fifteen minutes?

Assuming that the time between the arrival of two birds consecutive follows th exponential distribution and we need that this time exceeds fifteen minutes. If we convert the 15 minutes to hours we have 15(1/60)=0.25 hours. And we want to find this probability:

$P(T\geq 0.25h)$

And we can use the result obtained from the definitions and we have this:

$P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c. If you have already waited five minutes for the first bird to arrive, what is the probability that the bird will arrive within the next five minutes?

First we need to convert the 5 minutes to hours and we got 5(1/60)=0.0833h. And on this case we want a conditional probability. And for this case is good to remember the "Markovian property of the Exponential distribution", given by :

$P(T \leq a +t |T>t)=P(T\leq a)$

Since we have a waiting time for the first bird of 5 min = 0.0833h and we want that the next bird will arrive within 5 minutes=0.0833h, we can express on this way the probability of interest:

$P(T\leq 0.0833+0.0833| T>0.0833)$