Question

The number of spam emails received each day follows a Poisson
distribution with a mean of 50. Approximate the following
probabilities. Apply the ±½ correction factor and round value of
standard normal random variable to 2 decimal places.

Round your answer to four decimal places (e.g. 98.7654).

**(a)** More than 50 and less than 60 spam emails in a
day.

**(b)** At least 50 spam emails in a day.

**(c)** Less than 50 spam emails in a day.

**(d)** Approximate the probability that the total
number of spam emails exceeds 350 in a seven-day week.

Answer #1

here mean =50

and std deviation =sqrt(50)=7.0711

a)

for normal distribution z score =(X-μ)/σx | |

here mean= μ= | 50 |

std deviation =σ= | 7.0711 |

More than 50 and less than 60 spam emails in a day :

probability = | P(50.5<X<59.5) | = | P(0.07<Z<1.34)= | 0.9099-0.5279= | 0.3820 |

b)

At least 50 spam emails in a day :

probability = | P(X>49.5) | = | P(Z>-0.07)= | 1-P(Z<-0.07)= | 1-0.4721= | 0.5279 |

c)

Less than 50 spam emails in a day :

probability = | P(X<49.5) | = | P(Z<-0.07)= | 0.4721 |

d)

for 7 days; expected mail =50*7=350

and std deviation =sqrt(350)=18.7083

probability that the total number of spam emails exceeds 350 in a seven-day week :

probability = | P(X>350.5) | = | P(Z>0.027)= | 1-P(Z<0.03)= | 1-0.5120= | 0.4880 |

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