Probability : A certain area of USA is on the average hit by 6 hurricanes a year. Find the probability that in a given year this area will be hit by a) fewer than 4 hurricanes b) anywhere from 6 to 8 hurricanes
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if the average is 6 hurricanes, i believe that the answer is b. the probability that in a given year the area will be hit by anywhere from 6 to 8 hurricanes.
the problem is a poisson distribution with mean m = 6 \[P[x] = e^{-m *m^x/x!} \] a. \[P[x<4] = e^{-6(1+6 +6^2/2! +6^3/3!)} \]= ? <------ b. \[P[6≤x≤8] = e^{-6(6^6/6! + 6^7/7! + 6^8/8!)}\] = ?<------
hi r u there
yes?
if i write like that confirm plz
e-6*6^0 x e-6*6^1 x e-6*6^2 x e-6*6^3 and add-in all ans is it correct
you may want to rewrite it this way e^(-6*(1+6+(6^2)/2 + (6^3)/(3*2)))
Its B!!
Hi m still wondering how it comes.
fewer than 4 mean less than 4 i.e. 6^0 x 6^1 x 6^2 x 6^3
we use mew 6 and power four times of mew as i mentioned above?
\[P[x] = e ^{-m*m ^{x}/x!}\]
this is the probability distribution function for a Poison distribution. Whenever asked to find the probability of a Poison Distribution, you can use this formula. in this formula, m is the mean or expected value. and x is the variable. in this question, it is said that average hit of hurricanes is 6. which means people can expect an average of 6 hurricanes' hits each year.
like I said, as long as it's Poison Distribution, and asked for Probability. use this function. memorize this function.
i solved this question dear, and found the final answer is 0.13634. is it correct
I added four answers of equations
first one 6^0 second one 6^1 thrid one 6^2 fourth one 6^3
cuz x variables are 0 to 3 i.e. less than 4 or fewer than 4. correct me
\[P[x]= \frac{ m^x * e^{-m} }{ x! }\]
sorry i messed up the equation. this is the right one and yes you were right
x = {0, 1, 2, 3} and 4 is not included
i got answer i.e. 0.136349....is it correct
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