probability of acceptance a student to a collage number of Bernoulli trails

# Probability of acceptance a student to a collage

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(?) + 𝑷 ? (?) + 𝑷 ? (?) ? − probability of acceptance a student to a collage ? = ?, ? ? number of Bernoulli trails ? = ? ? students apply ? number of success in Bernoulli trails ? = ?, ?, ? 𝑷(?) = 𝑷(𝑿 ≤ ?) = 𝑷 ? (?) + 𝑷 ? (?) + 𝑷 ? (?) = probabilities from the table values in red (below) = ?, ???? + ?, ???? + ?, ???? = ?, ??? 𝑿 − a random variable has a binomial distribution 𝑿 ∶ ? ( ??, √??? ) so: µ = ?𝑿 = ?? = ? ∙ ?, ? = ?, ? it means that the most expected value of acceptance the students to a prestigious college is one and half student 𝑷 ? (?) Binomial Probabilities 𝑷 ? (?) = ( ? ? ) ∙ ? ? ∙ ? ?−? <fragment> n k 0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 5 0 0.95099 0.773781 0.59049 0.32768 0.16807 0.07776 0.03125 0.01024 0.00243 0.00032 1 0.04803 0.203627 0.32805 0.4096 0.36015 0.2592 0.15625 0.0768 0.02835 0.0064 2 0.00097 0.021434 0.0729 0.2048 0.3087 0.3456 0.3125 0.2304 0.1323 0.0512 3 9.8E-06 0.001128 0.0081 0.0512 0.1323 0.2304 0.3125 0.3456 0.3087 0.2048 4 4.95E-08 2.97E-05 0.00045 0.0064 0.02835 0.0768 0.15625 0.2592 0.36015 0.4096 5 1E-10 3.13E-07 0.00001 0.00032 0.00243 0.01024 0.03125 0.07776 0.16807 0.32768
Application of Mathematical Statistics (4) Iwona Nowakowska 5 Case Problem: In a certain company there were noticed that the average number of accidents in a month is 2 and the distribution of the number of accidents in this company can be described by Poisson distribution. Let 𝑿 − be a random variable describing the number of accidents in a month. 1) Find the probability that there will be no accidents in a randomly chosen month. 2) Find the probability that there will be 4 accidents in a randomly chosen month. 3) Find the probability that the number of accidents in a month will be at most 2. Solution: 𝑿 − the number of accidents in a month Let us notice that theoretically the random variable 𝑿 can assume any value 0,1,2,3, …. 𝑿 − has Poisson distribution, so we know that: 𝑿 ∶ 𝑷 ( ?, √? ) and: 𝑷(𝑿 = ?) = ? −? ? ? ?! where ? − the average number of events that occur during the given interval ? = ?, ?, ?, ?, … , ? − the number of successes 1) In our case: ? = ? , ? − no accidents ? = ? the average number of accidents in a month is 2
Application of Mathematical Statistics (4) Iwona Nowakowska 6 𝑷(𝑿 = ?) = 𝑷(𝑿 = ?) = ? −? ? ? ?! = ? −? ? ? ?! = ? −? ∙ ? = ? ? ? = = ? (?, ??) ? = ? ?, ???? = ?, ??? 2) In our case: ? = ? , ? − 4 accidents in a month ? = ? the average number of accidents in a month is 2 𝑷(𝑿 = ?) = 𝑷(𝑿 = ?) = ? −? ? ? ?! = ? −? ? ? ?! = ? −? ?? ?? = ? ? ? ? ? = = ? ? ? (?, ??) ? = ? ? ? ?, ???? = ? ??, ???? = ?, ?? 3) In our case: ? − number of accidents ? = ? or ? = ? or ? = ? ? = ? the average number of accidents in a month is 2 𝑷(𝑿 = ?) = ? −? ? ? ?! We look for the probability: 𝑷(𝑿 ≤ ?) = ? ? ? 𝑷(𝑿 ≤ ?) = 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) = ? ? ? 𝑷(𝑿 = ?) = ? −? ? ? ?! 𝑷(𝑿 ≤ ?) = 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) = = ? −? ? ? ?! + ? −? ? ? ?! + ? −? ? ? ?! =
Application of Mathematical Statistics (4) Iwona Nowakowska 7 = ? ? ? + ? ∙ ? ? ? + ? ? ? ? ? = ? ? ? + ? ∙ ? ? ? + ? ∙ ? ? ? = = ? ? ? = ? (?, ??) ? = ? ?, ???? = ?, ???? ≈ ?, ??? All above probabilities we can find in the table of Cumulative Poisson Probability Distribution. 1) 𝑷(𝑿 = ?) The probability 𝑷(𝑿 = ?) must be replaced by the probability 𝑷(𝑿 ≤ ?) the variable 𝑿 does not assume negative values. In the table we must find the value of the parameter ? = ? (in a column) and the value of ? = ? (in a row) We have immediately: 𝑷(𝑿 = ?) = ?, ???? value in red in the table (below) 2) 𝑷(𝑿 = ?) The probability 𝑷(𝑿 = ?) must be replaced by the probability 𝑷(𝑿 ≤ ?) − 𝑷(𝑿 ≤ ?) only such probabilities we have in the table We have immediately: 𝑷(𝑿 ≤ ?) − 𝑷(𝑿 ≤ ?) = ?, ???? − ?, ???? = ?, ???? ≈ ?, ??

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