5) Suppose that 14% of people are left handed. If 9 people are selected at random, what is the probability that exactly 2 of them are left handed? 6)
5) Suppose that 14% of people are left handed. If 9 people are selected at random, what is the probability that exactly 2 of them are left handed? 6)
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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![**Problem Statement:**
Write the formula you would use and set up how you would solve for the indicated probability (substitute in the given values into the formula). DO NOT USE A CALCULATOR/EXCEL ON THIS PROBLEM AND DO NOT SOLVE THE PROBLEM.
6) Suppose that 14% of people are left handed. If 9 people are selected at random, what is the probability that exactly 2 of them are left handed?
**Solution Setup:**
To find the probability of exactly 2 people being left handed out of 9, where the probability of a person being left handed is 14%, we use the Binomial Probability Formula:
\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \]
Where:
- \( n = 9 \) (total number of trials or people selected)
- \( k = 2 \) (number of successful outcomes we want, i.e., left-handed people)
- \( p = 0.14 \) (probability of a person being left handed)
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\)
Substitute the given values into the formula to set up the calculation.
\[ P(X = 2) = \binom{9}{2} \cdot (0.14)^2 \cdot (0.86)^{7} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fe73833-ff5a-4353-be1d-d2c3741e990b%2Ffa74aff9-427f-42dc-9c3b-5642d0a75eca%2Fyr7vbxl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Write the formula you would use and set up how you would solve for the indicated probability (substitute in the given values into the formula). DO NOT USE A CALCULATOR/EXCEL ON THIS PROBLEM AND DO NOT SOLVE THE PROBLEM.
6) Suppose that 14% of people are left handed. If 9 people are selected at random, what is the probability that exactly 2 of them are left handed?
**Solution Setup:**
To find the probability of exactly 2 people being left handed out of 9, where the probability of a person being left handed is 14%, we use the Binomial Probability Formula:
\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \]
Where:
- \( n = 9 \) (total number of trials or people selected)
- \( k = 2 \) (number of successful outcomes we want, i.e., left-handed people)
- \( p = 0.14 \) (probability of a person being left handed)
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\)
Substitute the given values into the formula to set up the calculation.
\[ P(X = 2) = \binom{9}{2} \cdot (0.14)^2 \cdot (0.86)^{7} \]
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