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Let x be a binomial random variable with n= 20 and p = 0.05. Calculate p(0) and p(1) using Table 1 to obtain the exact binomial probability. (Round your answers to three decimal places.) p(0) = 0.358 p(1) = 0.736 X

Let x be a binomial random variable with n= 20 and p = 0.05. Calculate p(0) and p(1) using Table 1 to obtain the exact binomial probability. (Round your answers to three decimal places.) p(0) = 0.358 p(1) = 0.736 X

MATLAB: An Introduction with Applications
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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Let \( x \) be a binomial random variable with \( n = 20 \) and \( p = 0.05 \). Calculate \( p(0) \) and \( p(1) \) using Table 1 to obtain the exact binomial probability. (Round your answers to three decimal places.) \[ p(0) = 0.358 \] ✅ \[ p(1) = 0.736 \] ❌
Transcribed Image Text:Let \( x \) be a binomial random variable with \( n = 20 \) and \( p = 0.05 \). Calculate \( p(0) \) and \( p(1) \) using Table 1 to obtain the exact binomial probability. (Round your answers to three decimal places.) \[ p(0) = 0.358 \] ✅ \[ p(1) = 0.736 \] ❌
**Table 1 (continued)** **Sample Size (n) = 20** This table provides probability values for different combinations of \( k \) and \( p \). The variables represented are: - \( k \) (ranging from 0 to 20): Represents the number of successes. - \( p \) (ranging from 0.01 to 0.99): Represents the probability of success on an individual trial. The table entries show the cumulative probability for a given \( k \) and \( p \). | \( k \) | .01 | .05 | .10 | .20 | .30 | .40 | .50 | .60 | .70 | .80 | .90 | .95 | .99 | \( k \) | |--------|------|------|------|------|------|------|------|------|------|------|------|------|------|--------| | 0 | .818 | .358 | .122 | .012 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 0 | | 1 | .983 | .736 | .392 | .069 | .008 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 1 | | 2 | .999 | .925 | .677 | .206 | .035 | .004 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 2 | | 3 | 1.000 | .984 | .867 | .411 | .107 | .016 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | 3 | | 4 | 1.000 | .997 | .957 | .630 | .238 | .051 | .006 | .000 | .000 | .000 | .000 | .000 | .000 | 4 | | 5 | 1.000 | 1.000 | .989 | .804 | .416 | .126 | .021 | .002 | .000 | .000 | .000 | .000 | .000 | 5 |
Transcribed Image Text:**Table 1 (continued)** **Sample Size (n) = 20** This table provides probability values for different combinations of \( k \) and \( p \). The variables represented are: - \( k \) (ranging from 0 to 20): Represents the number of successes. - \( p \) (ranging from 0.01 to 0.99): Represents the probability of success on an individual trial. The table entries show the cumulative probability for a given \( k \) and \( p \). | \( k \) | .01 | .05 | .10 | .20 | .30 | .40 | .50 | .60 | .70 | .80 | .90 | .95 | .99 | \( k \) | |--------|------|------|------|------|------|------|------|------|------|------|------|------|------|--------| | 0 | .818 | .358 | .122 | .012 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 0 | | 1 | .983 | .736 | .392 | .069 | .008 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 1 | | 2 | .999 | .925 | .677 | .206 | .035 | .004 | .000 | .000 | .000 | .000 | .000 | .000 | .000 | 2 | | 3 | 1.000 | .984 | .867 | .411 | .107 | .016 | .001 | .000 | .000 | .000 | .000 | .000 | .000 | 3 | | 4 | 1.000 | .997 | .957 | .630 | .238 | .051 | .006 | .000 | .000 | .000 | .000 | .000 | .000 | 4 | | 5 | 1.000 | 1.000 | .989 | .804 | .416 | .126 | .021 | .002 | .000 | .000 | .000 | .000 | .000 | 5 |
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