# 4.6 Poisson distribution  (Page 2/18)

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## Try it

An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution.

This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate. The events are independent.

## Notation for the poisson: p = poisson probability distribution function

X ~ P ( μ )

Read this as " X is a random variable with a Poisson distribution." The parameter is μ (or λ ); μ (or λ ) = the mean for the interval of interest.

Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?

Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $\frac{1}{4}$ hour.)

x = 0, 1, 2, 3, ...

If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives

$\left(\frac{1}{8}\right)$ (6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem.

X ~ P (0.75)

Find P ( x >1). P ( x >1) = 0.1734 (calculator or computer)

• Press 1 – and then press 2 nd DISTR.
• Arrow down to poissoncdf. Press ENTER.
• Enter (.75,1).
• The result is P ( x >1) = 0.1734.

## Note

The TI calculators use λ (lambda) for the mean.

The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:
P ( x >1) = 1 − poissoncdf(0.75, 1).

The graph of X ~ P (0.75) is:

The y -axis contains the probability of x where X = the number of calls in 15 minutes.

## Try it

A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer.

P ( x >4) = 0.0527

According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (147). The mean is 147 emails.

1. What is the probability that an email user receives exactly 160 emails per day?
2. What is the probability that an email user receives at most 160 emails per day?
3. What is the standard deviation?
1. P ( x = 160) = poissonpdf(147, 160) ≈ 0.0180
2. P ( x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666
3. Standard Deviation = $\sigma =\sqrt{\mu }=\sqrt{147}\approx 12.1244$

## Try it

According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (187). The mean is 187 text messages.

1. What is the probability that a teen girl sends exactly 175 texts per day?
2. What is the probability that a teen girl sends at most 150 texts per day?
3. What is the standard deviation?
1. P ( x = 175) = poissonpdf(187, 175) ≈ 0.0203
2. P ( x ≤ 150) = poissoncdf(187, 150) ≈ 0.0030
3. Standard Deviation =

7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
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Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
null rejected
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99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
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Niaz
what is null Hypothesis
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when median is greater than mode?
hello
Amaano
is this app useful
Worthy
little bit 😭
G-
oh
Worthy
when tail is positive
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define hypothesis
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I'm struggling to type it's on my laptop...statistics
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types of averages .mean median mode quarantiles MCQ question
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A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 7.6 and variance 0.01. Do you think that the mean thickness of the spearmint gum it produces is 7.5? (4 Points)
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10 gums mean = 7.6 variance= 0.01 standard deviation= ? what us the data set?
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0.6
Rubina