Consider the binomial distribution with n trials and AS) - p. (a) Show that Ply) Dly- 1) (n-v+ 1P for y- 1. 2.....n. Ecuivalently, for y- 1, 2, n, the equation pty) AEyP ply - 1) gives a recursive relationship between the probablities assodiated with successive values of Y. va Ply- 1) (," Jor - g-- 1) - n-y+ 1p (b) n- 70 and p- 0.04, use the above relationship to find PY < 3). (Round your answer to four decimal places.)
Consider the binomial distribution with n trials and AS) - p. (a) Show that Ply) Dly- 1) (n-v+ 1P for y- 1. 2.....n. Ecuivalently, for y- 1, 2, n, the equation pty) AEyP ply - 1) gives a recursive relationship between the probablities assodiated with successive values of Y. va Ply- 1) (," Jor - g-- 1) - n-y+ 1p (b) n- 70 and p- 0.04, use the above relationship to find PY < 3). (Round your answer to four decimal places.)
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:Consider the binomial distribution with n trials and PS) - p.
(a) Show that Ply)
Ply - 1)
(n - y+1)p
for y - 1, 2,..., n. Equivalently, for y- 1, 2, .., n, the equation plv) - M-Y+ 1)P ply - 1) gves a recursive relationship between the probabilities assodated with successive values of Y.
Ply - 1)
" pr-g- -1)
- In-y+ 1)p
(b) fn- 70 and p- 0.04, use the above relationship to find PY < 3). (Round your answer to four decimal places.)
x
Ply)
Ply - 1)
I-+1IP,1fy< (n + 1)p. This establishes that p(y) > ply - 1) if y is smalI (y < (n + 1)p) and p(y) < ply - 1) if y is large (y > (n + 1) p). Thus, successive binamial probabilities increase for a while and decrease from then on.
(c) Show that
ya
(n + 1)p > y
(n +1- vp >y+
(n +1- vo>
Your answer cannot be undenstood or graded. More Information
(n+1-yIP>
ya
plv)
Ply - 1)
<1 if y > (n + 1) p.
Show that
(n + 1lp <y
(n+1- vip < y -
(n +1- yiee
(n+1-ve
< 1
ya
plv)
Ply - 1)
(n + 1lp - y
Show that
-1 if (n + 1)p is an integer and y- (n + 1p.
(n+1- vip - y + (+1p
(n+1- vip -|F-1
(a+1-VIP - 1
ya
(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 + 1p - m for some linteger m, then p(m) - p(m - 1)-
Since for ys (n + 1)p, then ply) a
> Ply - 2) >..
Also for y 2 (n + 1p, then ply) a
> ply + 2) >.... Thus it is clear that p(y) is madimized when y is as clase to
pas possible.
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