By substitution, P(x) = (2) p*q" x P(x = 14) = (35) (0.397) ¹4 (0.603)35-14 = (35) (0.397) ¹4 (0.603)²¹ = 0.1365 Step 6: By substitution, P(x) = (n)p² qnx P(x < 17) = P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 16) P(x <17) = (35) (0.397)0 (0.603) 35-0+(35) (0.397)¹(0.603) 35-1 + (35) (0.397)2 (0.603) 35-2+...+(35) (0.397) ¹6 (0.603) 35-16 16 P (x < 17) = 0.8165 Step 7: By substitution, P(x) = (2) p* q-x + P(x = 12 13 P(x > 11) = P(x = 12) + P(x = 13) + P(x = 14) + - 35) P(x > 11) = (35) (0.397) ¹2 (0.603)35-12 + (35) (0.397) ¹3 (0.603) 35-13 + (35) (0.397) ¹4 (0.603) 35-14+...+(35) (0.397) ³5 (0.603) 35-35 35 P (x > 11) = 0.7947
By substitution, P(x) = (2) p*q" x P(x = 14) = (35) (0.397) ¹4 (0.603)35-14 = (35) (0.397) ¹4 (0.603)²¹ = 0.1365 Step 6: By substitution, P(x) = (n)p² qnx P(x < 17) = P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 16) P(x <17) = (35) (0.397)0 (0.603) 35-0+(35) (0.397)¹(0.603) 35-1 + (35) (0.397)2 (0.603) 35-2+...+(35) (0.397) ¹6 (0.603) 35-16 16 P (x < 17) = 0.8165 Step 7: By substitution, P(x) = (2) p* q-x + P(x = 12 13 P(x > 11) = P(x = 12) + P(x = 13) + P(x = 14) + - 35) P(x > 11) = (35) (0.397) ¹2 (0.603)35-12 + (35) (0.397) ¹3 (0.603) 35-13 + (35) (0.397) ¹4 (0.603) 35-14+...+(35) (0.397) ³5 (0.603) 35-35 35 P (x > 11) = 0.7947
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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I am needing help with understanding how we get the calculations for the powers.
If the equation is the same for the P=X... Why (P)X=14 has the power 14, but (P)X=17 and P=11? How do we get 0 in the probelm (P)X=17? why is it not 17?
Also in problem (P)X=11, why is X not 11 and the powers are 12 and 13. What makes these tow problems different from (P)X=14? How do we determine these other powers?
I hope my questions make sense..
See attached picture
Thank you so much!
Transcribed Image Text:By substitution,
P(x) = (2) p*q" x
P(x = 14) = (35) (0.397) ¹4 (0.603)35-14 = (35) (0.397) ¹4 (0.603)²¹ = 0.1365
Step 6:
By substitution,
P(x) = (n)p²
qnx
P(x < 17) = P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 16)
P(x <17) = (35) (0.397)0 (0.603) 35-0+(35) (0.397)¹(0.603) 35-1 +
(35) (0.397)2 (0.603) 35-2+...+(35) (0.397) ¹6 (0.603) 35-16
16
P (x < 17) = 0.8165
Step 7:
By substitution,
P(x) = (2) p* q-x
+ P(x =
12
13
P(x > 11) = P(x = 12) + P(x = 13) + P(x = 14) +
- 35)
P(x > 11) = (35) (0.397) ¹2 (0.603)35-12 + (35) (0.397) ¹3 (0.603) 35-13 +
(35) (0.397) ¹4 (0.603) 35-14+...+(35) (0.397) ³5 (0.603) 35-35
35
P (x > 11) = 0.7947
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