To address the problem, let's first calculate the probability distribution based on the given frequencies.
(a) Construct the probability distribution of X.
The probability \( P(X=x) \) can be calculated using the formula:
\[ P(X = x) = \frac{\text{Frequency of } x}{\text{Total Frequency}} = \frac{\text{Frequency of } x}{1682} \]
Let's calculate the probabilities for each number of hours:
| Number of Hours (x) | Frequency | P(X = x) | |----------------------|-----------|------------------------| | 0 | 113 | \( \frac{113}{1682} = 0.067 \) | | 1 | 184 | \( \frac{184}{1682} = 0.109 \) | | 2 | 336 | \( \frac{336}{1682} = 0.200 \) | | 3 | 251 | \( \frac{251}{1682} = 0.149 \) | | 4 | 318 | \( \frac{318}{1682} = 0.189 \) | | 5 | 234 | \( \frac{234}{1682} = 0.139 \) | | 6 | 151 | \( \frac{151}{1682} = 0.090 \) | | 7 | 35 | \( \frac{35}{1682} = 0.021 \) | | 8 | 60 | \( \frac{60}{1682} = 0.036 \) |
Thus, the probability distribution of \( X \) is:
\[ \begin{align*} P(X=0) & = 0.067 \ P(X=1) & = 0.109 \ P(X=2) & = 0.200 \ P(X=3) & = 0.149 \ P(X=4) & = 0.189 \ P(X=5) & = 0.139 \ P(X=6) & = 0.090 \ P(X=7) & = 0.021 \ P(X=8) & = 0.036 \ \end{align*} \]
(b) Find the probability that a person relaxes more than 5 hours per day.
To find \( P(X > 5) \):
\[ P(X > 5) = P(X=6) + P(X=7) + P(X=8) \]
Calculating this:
\[ P(X > 5) = 0.090 + 0.021 + 0.036 = 0.147 \]
(c) Find the probability that a person doesn't relax at all.
This is simply \( P(X = 0) \):
\[ P(X = 0) = 0.067 \]
(d) Compute the mean \( \mu_X \).
The mean \( \mu_X \) is calculated using:
\[ \mu_X = \sum (x \cdot P(X=x)) \]
Let's compute this:
\[ \begin{align*} \mu_X & = 0 \cdot 0.067 + 1 \cdot 0.109 + 2 \cdot 0.200 + 3 \cdot 0.149 + 4 \cdot 0.189 + 5 \cdot 0.139 + 6 \cdot 0.090 + 7 \cdot 0.021 + 8 \cdot 0.036 \ & = 0 + 0.109 + 0.400 + 0.447 + 0.756 + 0.695 + 0.540 + 0.147 + 0.288 \ & = 3.382 \end{align*} \]
Thus, the mean \( \mu_X \) rounded to two decimal places is:
\[ \mu_X \approx 3.38 \]
(e) Compute the standard deviation \( \sigma_X \).
First, calculate the variance \( \sigma^2_X \):
\[ \sigma^2_X = \sum (x^2 \cdot P(X=x)) - \mu_X^2 \]
To calculate this, we need \( x^2 \cdot P(X=x) \):
\[ \begin{align*} 0^2 \cdot 0.067 & = 0 \ 1^2 \cdot 0.109 & = 0.109 \ 2^2 \cdot 0.200 & = 0.800 \ 3^2 \cdot 0.149 & = 1.341 \ 4^2 \cdot 0.189 & = 3.024 \ 5^2 \cdot 0.139 & = 1.735 \ 6^2 \cdot 0.090 & = 3.240 \ 7^2 \cdot 0.021 & = 1.029 \ 8^2 \cdot 0.036 & = 2.304 \ \end{align*} \]
Now sum these products:
\[ 0 + 0.109 + 0.800 + 1.341 + 3.024 + 1.735 + 3.240 + 1.029 + 2.304 = 13.582 \]
Now, plug this into the variance formula:
\[ \sigma^2_X = 13.582 - (3.382)^2 = 13.582 - 11.438 = 2.144 \]
Finally, calculate the standard deviation:
\[ \sigma_X = \sqrt{2.144} \approx 1.464 \]
Thus, the standard deviation \( \sigma_X \) rounded to three decimal places is:
\[ \sigma_X \approx 1.464 \]
Summary of Answers:
(a) Probability distribution: given above.
(b) \( P(X > 5) = 0.147 \)
(c) \( P(X = 0) = 0.067 \)
(d) Mean \( \mu_X \approx 3.38 \)
(e) Standard deviation \( \sigma_X \approx 1.464 \)