Question is not well presented
Consider the binomial distribution with n trials and P(S) = p.
Show that
p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.
This establishes that p(y) > p(y − 1) if y is small (y < (n + 1)p) and p(y) < p(y − 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on.
Answer:
See Explanation Below
Step-by-step explanation:
Given that
p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.
First, we make the following assumption
p(y) /p(y − 1) = (n − y + 1)p /yq = 1;
So, we have
(n − y + 1)p /yq = 1
(n − y + 1)p = yq
Note that p + q = 1;.
So, q = 1 - p
Substitute 1-p for q in the above expression
(n − y + 1)p = y(1-p)
np - py + p = y - py
Solve for y
..... Collect like terms
np + p = y - py + py
np + p = y;
So, y = np + p
y = (n + 1)p
From the above,
We know that
p(y) /p(y − 1) = 1
if y = (n + 1)p.
Similarly, we've also obtain that
p(y) /p(y − 1) > 1 if y < (n + 1)p.
