(e) Show that PY) ply - 1) G-y+1)P > 1 if y < (n + 1)p. This establishes that ply) > plY - 1) if y is small y < (n - 1)p) and ply) < Ply - 1) # y is large (y > (n - 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. (n - 1)p > y (n +1- y)p > y n+1 (n +1- y)p > (n +1- ylp Show that PY <1 if y > (n + 1) p. Py - 1) (n + 1)p ply-2) >... Also for y 2 (n + 1)p, then p(y) zA > p(y + 2) >. Thus it is clear that p(y) is maximized when y is as close to p as possible.
(e) Show that PY) ply - 1) G-y+1)P > 1 if y < (n + 1)p. This establishes that ply) > plY - 1) if y is small y < (n - 1)p) and ply) < Ply - 1) # y is large (y > (n - 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. (n - 1)p > y (n +1- y)p > y n+1 (n +1- y)p > (n +1- ylp Show that PY <1 if y > (n + 1) p. Py - 1) (n + 1)p ply-2) >... Also for y 2 (n + 1)p, then p(y) zA > p(y + 2) >. Thus it is clear that p(y) is maximized when y is as close to p as possible.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:() Show that PM) = -y + 1)P> 1 if y < (n + 1)p. This establishes that p(y) > P(y - 1) if y is small (y < (n + 1)p) and ply) < p(v - 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on.
Ply -
(n + 1)p > y
yg
(n +1- y)p > y + n+1
(n +1- y)p >
(n +1- y)P > 1
yq
Show that_P)
PŮY – 1)
(n + 1)p < y
<1 if y > (n + 1) p.
(n +1- y)p < y-
(n +1- y)p <
(n +1- y)p
< 1
yg
Show that P)
Ply - 1)
= 1 if (n + 1)p is an integer and y = (n + 1)p.
(n + 1)p = y
(n +1- y)p = y + (n+ 1)p
(n +1- y)p = Y- 1
(n +1- y)p = 1
yq
(d) Show that the value of y assigned the largest probability is equal to the greatest integer less than or equal to (n + 1)p. If (n + 1)p = m for some integer m, then p(m) = p(m - 1).
Since for y s (n + 1)p, then p(y) z pl
> Ply - 2) >
Also for y 2 (n + 1)p, then p(y) z
> P(y + 2) >
. Thus it is clear that p(y) is maximized when y is as close to
p as possible.
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