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.

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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.
Transcribed Image Text:**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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