Mathematics, 07.03.2020 03:44 jasminemora4

The number of emails that I get in a weekday (Monday through Friday) can be modeled by a Poisson distribution with an average of 1/6 emails per minute. The number of emails that I receive on weekends (Saturday and Sunday) can be modeled by a Poisson distribution with an average of 1/30 emails per minute.

What is the probability that I get no emails in an interval of length 4 hours on a Sunday?

A random day is chosen (all days of the week are equally likely to be selected), and a random interval of length one hour is selected on the chosen day. It is observed that I did not receive any emails in that interval.

What is the probability that the chosen day is a weekday?

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Answer from: andrejr0330jr

1) P(x=0) = 0.0003355

2) P(U|W) = 0.000838

Step-by-step explanation:

The number of emails received on Sunday can be modeled by a Poisson distribution with an average of 1/30 emails per minute.

Also; taking the probability that no email in an interval of the length of 4 hours on Sunday; we have:

Time (T) = 4 hours

= 4 × 60 minutes

= 240 minutes

Poisson Distribution can be expressed as:

$\lambda = T * \frac{1}{30}$

$\lambda=$ $240*\frac{1}{30}$

$\lambda=$ 8

The Poisson Probability can now be calculated by using the formula;

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

$P(x=0) = \frac{e^ {8} \lambda^0}{0!}$

P(x=0) = $e^{-8}$

= 0.0003355

A random day is chosen (all days of the week are equally likely to be selected), and a random interval of length one hour is selected on the chosen day. It is observed that I did not receive any emails in that interval.

What is the probability that the chosen day is a weekday?

Let use U to represent the event that weekday(Monday - Friday) is chosen.

Hence, P(U) = $\frac{5}{7}$

Let V represent the event that weekend( Saturday or Sunday ) is chosen.

P(V) = $\frac{2}{7}$

Let W represent the event that no emails were received.

The probability that no emails were received during the weekday can then be illustrated as:

P ( W║U)

The Poisson Distribution for weekday (Monday - Friday ) is given by:

$\lambda = \frac{1}{6} *1 hour$

$\lambda = \frac{60}{6}minutes$

$\lambda = 10 minutes$

Probability P ( W║U) for Poisson Probability can now be determined as:

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

$P(x=0) = \frac{e^ {-10} \lambda^0}{0!}$

P(x=0) = $e^{-10}$

Probability that no emails we received during the weekend (Saturday or Sunday) can be illustrated as: P ( W║V).

The Poisson distribution for weekend (Saturday or Sunday) is given by:

$\lambda = \frac{1}{30} *1 hour$

$\lambda = \frac{60}{30}minutes$

$\lambda = 2 minutes$

P ( W║V) for Poisson Probability can be be calculated as:

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

$P(x=0) = \frac{e^ {-2} \lambda^0}{0!}$

P(x=0) = $e^{-2}$

FInally, the probability that the chosen day is a weekday given that no emails were received in the interval of the length one hour can now be determined as:

$P ( U|W) =\frac{P(W|U)P(U)}{P(W)}$

$P ( U|W) =\frac{P(W|U)P(U)}{P(W|U)P(U)+P(W|V)P(V)}$

P(U|W) = $\frac{e^{-10}*\frac{5}{7} }{(e^{-10}*\frac{5}{7})+(e^{-2}*\frac{2}{7}) }$

P(U|W) = $\frac{5e^{-10}}{5e^{-10}+2e^{-2}}$

P(U|W) = $\frac{5e^{-10}}{e^{-10}(2+e^{8})}$

P(U|W) = $\frac{5}{(5+2e^8)}$

P(U|W) = 0.000838

Answer from: Quest

m a t h w a y. c o m

step-by-step explanation:

Answer from: Quest

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

In the diagram of circle o, what is the measure of abc? 210 150

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The number of emails that I get in a weekday (Monday through Friday) can be modeled by a Poisson dis...