**Probability Distribution and Calculations for Garage Spaces**The probability that a randomly selected home has a garage space for 0, 1, 2, or 3 cars is 0.22, 0.33, 0.37, and 0.08, respectively.### a) Probability Distribution Table| Garage Space (X) | P(X) ||------------------|------|| 0 | 0.22 || 1 | 0.33 || 2 | 0.37 || 3 | 0.08 |This table outlines the probability distribution for garage spaces available in a randomly selected home.### b) Calculations**Expected Value (E(X)), Population Mean (μ):**- Calculated as the sum of each value multiplied by its probability: \[ \mu = E(X) = (0 \times 0.22) + (1 \times 0.33) + (2 \times 0.37) + (3 \times 0.08) = 1.31 \]**Population Standard Deviation (σ):**- Square root of the variance (yet to be calculated).**Population Variance (σ²):**To be calculated and then used to find the standard deviation.### c) Interpretation of Population MeanThe population mean of 1.31 indicates that, on average, a home is expected to have about 1.31 garage spaces when many homes are considered.### d) Interpretation of Population Standard DeviationThe population standard deviation, approximately 0.90 (calculation not shown here), indicates the variability in the number of garage spaces across different homes.

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Description: **Probability Distribution and Calculations for Garage Spaces**The probability that a randomly selected home has a garage space for 0, 1, 2, or 3 cars is 0.22, 0.33, 0.37, and 0.08, respectively.### a) Probability Distribution Table| Garage Space (X

(a) Construct the probability distribution table. The probability distribution table is constructed…

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The probability that a randomly selected home has a garage space for 0, 1, 2, 3 cars are 0.22, 0.33, 0.37, and 0.08, respectively. a) Construct a probability distribution for the data. Garage Space, X ♥ 1. P(x) 0.22 0.33 0.37 0.08 b) Calculate E(X) (same as µ), o², and o. Round to two decimal places. Population Mean: u マ = 1.31 Population Standard Deviation: o Population Variance: 0 %3D c) What does the population mean tell us about this problem? If we look at many homes, we expect to see on average about 1.31 garage spaces. d) What does the population standard deviation tell us about this problem? If we look at many homes, we expect the variation to be about 0.90 garage spaces.

The probability that a randomly selected home has a garage space for 0, 1, 2, 3 cars are 0.22, 0.33, 0.37, and 0.08, respectively. a) Construct a probability distribution for the data. Garage Space, X ♥ 1. P(x) 0.22 0.33 0.37 0.08 b) Calculate E(X) (same as µ), o², and o. Round to two decimal places. Population Mean: u マ = 1.31 Population Standard Deviation: o Population Variance: 0 %3D c) What does the population mean tell us about this problem? If we look at many homes, we expect to see on average about 1.31 garage spaces. d) What does the population standard deviation tell us about this problem? If we look at many homes, we expect the variation to be about 0.90 garage spaces.

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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Concept explainers

Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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**Probability Distribution and Calculations for Garage Spaces**

The probability that a randomly selected home has a garage space for 0, 1, 2, or 3 cars is 0.22, 0.33, 0.37, and 0.08, respectively.

### a) Probability Distribution Table

| Garage Space (X) | P(X) |
|------------------|------|
| 0 | 0.22 |
| 1 | 0.33 |
| 2 | 0.37 |
| 3 | 0.08 |

This table outlines the probability distribution for garage spaces available in a randomly selected home.

### b) Calculations

**Expected Value (E(X)), Population Mean (μ):**

- Calculated as the sum of each value multiplied by its probability:
\[
\mu = E(X) = (0 \times 0.22) + (1 \times 0.33) + (2 \times 0.37) + (3 \times 0.08) = 1.31
\]

**Population Standard Deviation (σ):**

- Square root of the variance (yet to be calculated).

**Population Variance (σ²):**

To be calculated and then used to find the standard deviation.

### c) Interpretation of Population Mean

The population mean of 1.31 indicates that, on average, a home is expected to have about 1.31 garage spaces when many homes are considered.

### d) Interpretation of Population Standard Deviation

The population standard deviation, approximately 0.90 (calculation not shown here), indicates the variability in the number of garage spaces across different homes.

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