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### Probability Distribution of Pizza Toppings In a pizza takeout restaurant, a probability distribution was obtained for the number of toppings on a large pizza. The random variable \( x \) represents the number of toppings. The task is to complete the table and use the data to find the mean and standard deviation. #### Probability Distribution Table | \( x \) | \( P(x) \) | \( xP(x) \) | \( (x - \mu) \) | \( (x - \mu)^2 \) | \( (x - \mu)^2 P(x) \) | |---------|------------|-------------|-----------------|-------------------|-----------------------| | 0 | 0.36 | 0 | -1.86 | 3.4596 | 1.245456 | | 1 | 0.08 | 0.08 | -0.86 | 0.7396 | 0.059168 | | 2 | 0.20 | 0.4 | 0.14 | 0.0196 | 0.00392 | | 3 | 0.06 | | | | | | 4 | 0.30 | | | | | To complete the table, calculate \( xP(x) \), \( (x - \mu) \), \( (x - \mu)^2 \), and \( (x - \mu)^2 P(x) \) for each value of \( x \). ### Mean and Standard Deviation Options: A. mean: 0.48; standard deviation: 1.66 B. mean: 1.86; standard deviation: 1.66 C. mean: 0.48; standard deviation: 2.76 D. mean: 1.86; standard deviation: 2.76 E. mean: 1.86; standard deviation: 1.31 ### Additional Problem: Binomial Probability An airline reports a 20% rate of no-shows on advanced reservations. With 230 reservations, find the probability of exactly 42 no-shows. Use continuity correction and the standard normal distribution table to approximate the binomial probability.

### Probability Distribution of Pizza Toppings In a pizza takeout restaurant, a probability distribution was obtained for the number of toppings on a large pizza. The random variable \( x \) represents the number of toppings. The task is to complete the table and use the data to find the mean and standard deviation. #### Probability Distribution Table | \( x \) | \( P(x) \) | \( xP(x) \) | \( (x - \mu) \) | \( (x - \mu)^2 \) | \( (x - \mu)^2 P(x) \) | |---------|------------|-------------|-----------------|-------------------|-----------------------| | 0 | 0.36 | 0 | -1.86 | 3.4596 | 1.245456 | | 1 | 0.08 | 0.08 | -0.86 | 0.7396 | 0.059168 | | 2 | 0.20 | 0.4 | 0.14 | 0.0196 | 0.00392 | | 3 | 0.06 | | | | | | 4 | 0.30 | | | | | To complete the table, calculate \( xP(x) \), \( (x - \mu) \), \( (x - \mu)^2 \), and \( (x - \mu)^2 P(x) \) for each value of \( x \). ### Mean and Standard Deviation Options: A. mean: 0.48; standard deviation: 1.66 B. mean: 1.86; standard deviation: 1.66 C. mean: 0.48; standard deviation: 2.76 D. mean: 1.86; standard deviation: 2.76 E. mean: 1.86; standard deviation: 1.31 ### Additional Problem: Binomial Probability An airline reports a 20% rate of no-shows on advanced reservations. With 230 reservations, find the probability of exactly 42 no-shows. Use continuity correction and the standard normal distribution table to approximate the binomial probability.

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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### Probability Distribution of Pizza Toppings In a pizza takeout restaurant, a probability distribution was obtained for the number of toppings on a large pizza. The random variable \( x \) represents the number of toppings. The task is to complete the table and use the data to find the mean and standard deviation. #### Probability Distribution Table | \( x \) | \( P(x) \) | \( xP(x) \) | \( (x - \mu) \) | \( (x - \mu)^2 \) | \( (x - \mu)^2 P(x) \) | |---------|------------|-------------|-----------------|-------------------|-----------------------| | 0 | 0.36 | 0 | -1.86 | 3.4596 | 1.245456 | | 1 | 0.08 | 0.08 | -0.86 | 0.7396 | 0.059168 | | 2 | 0.20 | 0.4 | 0.14 | 0.0196 | 0.00392 | | 3 | 0.06 | | | | | | 4 | 0.30 | | | | | To complete the table, calculate \( xP(x) \), \( (x - \mu) \), \( (x - \mu)^2 \), and \( (x - \mu)^2 P(x) \) for each value of \( x \). ### Mean and Standard Deviation Options: A. mean: 0.48; standard deviation: 1.66 B. mean: 1.86; standard deviation: 1.66 C. mean: 0.48; standard deviation: 2.76 D. mean: 1.86; standard deviation: 2.76 E. mean: 1.86; standard deviation: 1.31 ### Additional Problem: Binomial Probability An airline reports a 20% rate of no-shows on advanced reservations. With 230 reservations, find the probability of exactly 42 no-shows. Use continuity correction and the standard normal distribution table to approximate the binomial probability.
Transcribed Image Text:### Probability Distribution of Pizza Toppings In a pizza takeout restaurant, a probability distribution was obtained for the number of toppings on a large pizza. The random variable \( x \) represents the number of toppings. The task is to complete the table and use the data to find the mean and standard deviation. #### Probability Distribution Table | \( x \) | \( P(x) \) | \( xP(x) \) | \( (x - \mu) \) | \( (x - \mu)^2 \) | \( (x - \mu)^2 P(x) \) | |---------|------------|-------------|-----------------|-------------------|-----------------------| | 0 | 0.36 | 0 | -1.86 | 3.4596 | 1.245456 | | 1 | 0.08 | 0.08 | -0.86 | 0.7396 | 0.059168 | | 2 | 0.20 | 0.4 | 0.14 | 0.0196 | 0.00392 | | 3 | 0.06 | | | | | | 4 | 0.30 | | | | | To complete the table, calculate \( xP(x) \), \( (x - \mu) \), \( (x - \mu)^2 \), and \( (x - \mu)^2 P(x) \) for each value of \( x \). ### Mean and Standard Deviation Options: A. mean: 0.48; standard deviation: 1.66 B. mean: 1.86; standard deviation: 1.66 C. mean: 0.48; standard deviation: 2.76 D. mean: 1.86; standard deviation: 2.76 E. mean: 1.86; standard deviation: 1.31 ### Additional Problem: Binomial Probability An airline reports a 20% rate of no-shows on advanced reservations. With 230 reservations, find the probability of exactly 42 no-shows. Use continuity correction and the standard normal distribution table to approximate the binomial probability.
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