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yashar806
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A random variable X has a binomial distribution with parameters n = 4 and p. What must be the value of the parameter p if we know that P(X = 3)= P(X = 4)?
 one year ago
 one year ago
yashar806 Group Title
A random variable X has a binomial distribution with parameters n = 4 and p. What must be the value of the parameter p if we know that P(X = 3)= P(X = 4)?
 one year ago
 one year ago

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yashar806 Group TitleBest ResponseYou've already chosen the best response.0
can you help?
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
please help
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
\[P(X=3)=\left(\begin{matrix}4 \\ 3\end{matrix}\right)p ^{3}(1p)=4p ^{3}(1p)=4p ^{3}4p ^{4}\] \[P(X=4)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)p ^{4}(1p)^{0}=p ^{4}\] \[4p ^{3}4p ^{4}=p ^{4}\ ....................(1)\] Now solve equation (1) to find the value of p.
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
i cant solve that euqation?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Do you want me to help you to solve the equation?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
\[4p ^{3}4p ^{4}=p ^{4}\] First add 4p^4 to both sides.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
What did you get?
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
4p^3= 5p^4
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Now divide both sides by 5p^3 and post the result.
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
i dont get it
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{4p ^{3}}{5p ^{3}}=\frac{5p ^{4}}{5p ^{3}}\] Now simplify.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Yes, p = 4/5.
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
but the answer is 3/4
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
If you test the value of p = 3/4 you will find that it is not correct. Testing the value of p = 4/5 shows that it is correct. Did you copy the original question correctly?
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
yes i did
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
anyway, i gotthe idea, can I ask you one more question, i just need an explanation, please
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
A variable X follows a normal distribution with mean 10 and standard deviation 5. Another variable Y follows a normal distribution with mean 25 and standard deviation 10. The maximum height of the density curve for X is_______ (i) the maximum height for the density curve for Y , and the area under the density curve for X is__________ (ii) the area under the density curve for Y . (A) (i) greater than, (ii) less than (B) (i) less than, (ii) greater than (C) (i) equal to, (ii) equal to (D) (i) greater than, (ii) equal to (E) (i) less than, (ii) less than
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Here is the checking of the answer p = 4/5: \[P(X=3)=\left(\begin{matrix}4 \\ 3\end{matrix}\right)\times (\frac{4}{5})^{3}\times \frac{1}{5}=(\frac{4}{5})^{4}\] \[P(X=4)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)\times (\frac{4}{5})^{4}\times (\frac{1}{5})^{0}=(\frac{4}{5})^{4}\]
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
could you explain that one please
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
@yashar806 Please post your second question again after closing this one. Sorry I cannot help any further. I must log out now.
 one year ago
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