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.
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.
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.
What is the probability that an email user receives exactly 160 emails per day?
What is the probability that an email user receives at most 160 emails per day?
What is the standard deviation?
P (
x = 160) = poissonpdf(147, 160) ≈ 0.0180
P (
x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666
Standard Deviation =
$\sigma =\sqrt{\mu}=\sqrt{147}\approx 12.1244$
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.
What is the probability that a teen girl sends exactly 175 texts per day?
What is the probability that a teen girl sends at most 150 texts per day?
What is the standard deviation?
P (
x = 175) = poissonpdf(187, 175) ≈ 0.0203
P (
x ≤ 150) = poissoncdf(187, 150) ≈ 0.0030
Standard Deviation =
$\sigma =\sqrt{\mu}\text{=}\sqrt{187}\approx 13.6748$
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.
a bad contain 3 red and 5 black balls another 4 red and 7 black balls, A ball is drawn from a bag selected at random, Find the probability that A is red?
The information is given as, 30% of customers shopping at SHOPNO will switch to DAILY SHOPPING every month on the other hand 40% of customers shopping at DAILY SHOPPING will switch to other every month. What is the probability that customers will switch from A to B for next two months?
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
3. The following are the number of mails received in different days by different organizations:
Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67.
Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25.
i) Fit a regression line of y on x and test the significance of regression.
ii) Estimate y
The number of problem creating computers of two laboratories are as follows:
Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12.
Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38.
Are the two laboratories similar in respect of problem creating compute
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?
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 standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
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?
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)
Omer
10 gums
mean = 7.6
variance= 0.01
standard deviation= ?
what us the data set?