snurpeisov

07.10.2020 •

Mathematics

Suppose that the following frequency table contains data on the number of absences in a school year due to illness or injury for 100 randomly selected high school students. Use the table to determine the median number of absences for these 100 students. Provide your answer with precision to one decimal place. Number of absences 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Frequency 25, 10, 10, 5, 9, 12, 9, 9, 4, 2, 3, 2

Solved

Show answers

Ответ:

Reward18

07.01.2021

Mathematics

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X 2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X 2.725 + 2*5.540) = P(X13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X 2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X 2.725 + 2*5.540) = P(X13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

I want advice