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Estimate P(6) for n = 18 and p= 0.3 by using the normal distribution as an approximation to the binomial distribution. Round to four decimal places. ..... O A. 0.1239 B. 0.1937 O C. 0.1015 O D. 0.8513

Estimate P(6) for n = 18 and p= 0.3 by using the normal distribution as an approximation to the binomial distribution. Round to four decimal places. ..... O A. 0.1239 B. 0.1937 O C. 0.1015 O D. 0.8513

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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I was able to solve this by guessing. Please show me how to do the work and use excel if you can. Thanks!

**Problem:** Estimate \( P(6) \) for \( n = 18 \) and \( p = 0.3 \) by using the normal distribution as an approximation to the binomial distribution. Round to four decimal places. **Options:** - A. 0.1239 - **B. 0.1937** (correct answer) - C. 0.1015 - D. 0.8513 **Explanation:** To approximate a binomial probability using the normal distribution, we use the normal approximation for the binomial distribution given by: 1. Calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)): - \(\mu = n \times p\) - \(\sigma = \sqrt{n \times p \times (1-p)}\) 2. Convert the binomial probability to a z-score using \( X = 6 \): - \( z = \frac{X - \mu}{\sigma} \) 3. Use the z-score to find the corresponding probability from the standard normal distribution table. The answer, rounded to four decimal places, is 0.1937, corresponding to option B.
Transcribed Image Text:**Problem:** Estimate \( P(6) \) for \( n = 18 \) and \( p = 0.3 \) by using the normal distribution as an approximation to the binomial distribution. Round to four decimal places. **Options:** - A. 0.1239 - **B. 0.1937** (correct answer) - C. 0.1015 - D. 0.8513 **Explanation:** To approximate a binomial probability using the normal distribution, we use the normal approximation for the binomial distribution given by: 1. Calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)): - \(\mu = n \times p\) - \(\sigma = \sqrt{n \times p \times (1-p)}\) 2. Convert the binomial probability to a z-score using \( X = 6 \): - \( z = \frac{X - \mu}{\sigma} \) 3. Use the z-score to find the corresponding probability from the standard normal distribution table. The answer, rounded to four decimal places, is 0.1937, corresponding to option B.
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