A recent survey asked 1677 people how many hours per day they were able to relax. The results are presented in the following table.

Question

Number of hours: 0 1 2 3 4 5 6 7 8
Frequency: 116, 187, 336, 251, 315, 229, 150, 31, 62 with a total of 1677
Consider these 1677 people to be a population. Let X be the number of hours of relaxation for a person sampled at random from this population.
(a) Construct the probability distribution of X. Round the answer to three decimal places.
(b) Find the probability that a person relaxes more than 5 hours per day.
(c) Find the probability that a person doesn't relax at all.
d) Compute the mean UvX. Round the answer to two decimal places.
(e) Compute the standard deviation OvX. Round the answer to three decimal places.

Answer 1

To answer your questions, let's first structure the data given and compute the required values step-by-step.

(a) Construct the probability distribution of X.

The probability of a certain number of hours of relaxation can be calculated for each category by dividing the frequency of that category by the total number of respondents.

Here’s the data:

• Total respondents \( N = 1677 \)

| Hours of Relaxation (X) | Frequency (f) | Probability P(X) | |-------------------------|---------------|---------------------------| | 0 | 116 | \( P(0) = \frac{116}{1677} \) | | 1 | 187 | \( P(1) = \frac{187}{1677} \) | | 2 | 336 | \( P(2) = \frac{336}{1677} \) | | 3 | 251 | \( P(3) = \frac{251}{1677} \) | | 4 | 315 | \( P(4) = \frac{315}{1677} \) | | 5 | 229 | \( P(5) = \frac{229}{1677} \) | | 6 | 150 | \( P(6) = \frac{150}{1677} \) | | 7 | 31 | \( P(7) = \frac{31}{1677} \) | | 8 | 62 | \( P(8) = \frac{62}{1677} \) |

Now, let's calculate the probabilities:

1. \( P(0) = \frac{116}{1677} \approx 0.069 \)
2. \( P(1) = \frac{187}{1677} \approx 0.111 \)
3. \( P(2) = \frac{336}{1677} \approx 0.200 \)
4. \( P(3) = \frac{251}{1677} \approx 0.150 \)
5. \( P(4) = \frac{315}{1677} \approx 0.188 \)
6. \( P(5) = \frac{229}{1677} \approx 0.136 \)
7. \( P(6) = \frac{150}{1677} \approx 0.089 \)
8. \( P(7) = \frac{31}{1677} \approx 0.018 \)
9. \( P(8) = \frac{62}{1677} \approx 0.037 \)

Now structuring this in a table:

| Hours | Probability (P(X)) | |--------|---------------------| | 0 | 0.069 | | 1 | 0.111 | | 2 | 0.200 | | 3 | 0.150 | | 4 | 0.188 | | 5 | 0.136 | | 6 | 0.089 | | 7 | 0.018 | | 8 | 0.037 |

(b) Find the probability that a person relaxes more than 5 hours per day.

To find \( P(X > 5) \), we sum the probabilities of 6, 7, and 8 hours:

\[ P(X > 5) = P(6) + P(7) + P(8) \ = 0.089 + 0.018 + 0.037 = 0.144 \]

(c) Find the probability that a person doesn't relax at all.

From part (a): \[ P(X = 0) = P(0) = 0.069 \]

(d) Compute the mean \( \mu_X \).

The mean (expected value) can be calculated as follows:

\[ \mu_X = \sum (X_i \cdot P(X_i)) \]

Calculating this:

\[ \mu_X = (0 \cdot 0.069) + (1 \cdot 0.111) + (2 \cdot 0.200) + (3 \cdot 0.150) + (4 \cdot 0.188) + (5 \cdot 0.136) + (6 \cdot 0.089) + (7 \cdot 0.018) + (8 \cdot 0.037) \]

Calculating each term:

• \( 0 \cdot 0.069 = 0 \)
• \( 1 \cdot 0.111 = 0.111 \)
• \( 2 \cdot 0.200 = 0.400 \)
• \( 3 \cdot 0.150 = 0.450 \)
• \( 4 \cdot 0.188 = 0.752 \)
• \( 5 \cdot 0.136 = 0.680 \)
• \( 6 \cdot 0.089 = 0.534 \)
• \( 7 \cdot 0.018 = 0.126 \)
• \( 8 \cdot 0.037 = 0.296 \)

Now summing these values:

\[ \mu_X = 0 + 0.111 + 0.400 + 0.450 + 0.752 + 0.680 + 0.534 + 0.126 + 0.296 = 3.407 \]

(e) Compute the standard deviation \( \sigma_X \).

First, we need to calculate the variance \( \sigma_X^2 \).

\[ \sigma_X^2 = \sum (X_i^2 \cdot P(X_i)) - \mu_X^2 \]

Calculating \( \sum (X_i^2 \cdot P(X_i)) \):

\[ = (0^2 \cdot 0.069) + (1^2 \cdot 0.111) + (2^2 \cdot 0.200) + (3^2 \cdot 0.150) + (4^2 \cdot 0.188) + (5^2 \cdot 0.136) + (6^2 \cdot 0.089) + (7^2 \cdot 0.018) + (8^2 \cdot 0.037) \]

Calculating each term:

• \( 0^2 \cdot 0.069 = 0 \)
• \( 1^2 \cdot 0.111 = 0.111 \)
• \( 2^2 \cdot 0.200 = 0.800 \)
• \( 3^2 \cdot 0.150 = 1.350 \)
• \( 4^2 \cdot 0.188 = 3.008 \)
• \( 5^2 \cdot 0.136 = 1.700 \)
• \( 6^2 \cdot 0.089 = 3.204 \)
• \( 7^2 \cdot 0.018 = 0.882 \)
• \( 8^2 \cdot 0.037 = 0.592 \)

Now summing these values:

\[ = 0 + 0.111 + 0.800 + 1.350 + 3.008 + 1.700 + 3.204 + 0.882 + 0.592 = 11.647 \]

Now applying the variance formula:

\[ \sigma_X^2 = 11.647 - (3.407)^2 = 11.647 - 11.603649 = 0.043351 \]

Finally, calculate the standard deviation:

\[ \sigma_X = \sqrt{0.043351} \approx 0.208 \]

Summary of Results

(a) Probability Distribution:

| Hours | Probability (P(X)) | |-------|---------------------| | 0 | 0.069 | | 1 | 0.111 | | 2 | 0.200 | | 3 | 0.150 | | 4 | 0.188 | | 5 | 0.136 | | 6 | 0.089 | | 7 | 0.018 | | 8 | 0.037 |

(b) \( P(X > 5) = 0.144 \)

(c) \( P(X = 0) = 0.069 \)

(d) Mean \( \mu_X = 3.41 \)

(e) Standard Deviation \( \sigma_X \approx 0.208 \)

Answer 2

The answer for 1 hour is wrong, The answer for 5 hours is wrong.
D and E are wrong.
Everything else is correct.

Answer 3

Let's carefully recalculate the probabilities for 1 hour and 5 hours, as well as recompute the mean and standard deviation based on these corrections.

(a) Probability Distribution of X:

First, let's verify the entire probability distribution one more time.

Total respondents \( N = 1677 \).

| Hours of Relaxation (X) | Frequency (f) | Probability P(X) | |-------------------------|---------------|---------------------------| | 0 | 116 | \( P(0) = \frac{116}{1677} \approx 0.069 \) | | 1 | 187 | \( P(1) = \frac{187}{1677} \approx 0.111 \) | | 2 | 336 | \( P(2) = \frac{336}{1677} \approx 0.200 \) | | 3 | 251 | \( P(3) = \frac{251}{1677} \approx 0.150 \) | | 4 | 315 | \( P(4) = \frac{315}{1677} \approx 0.188 \) | | 5 | 229 | \( P(5) = \frac{229}{1677} \approx 0.136 \) | | 6 | 150 | \( P(6) = \frac{150}{1677} \approx 0.089 \) | | 7 | 31 | \( P(7) = \frac{31}{1677} \approx 0.018 \) | | 8 | 62 | \( P(8) = \frac{62}{1677} \approx 0.037 \) |

So the probabilities are correct as shown, including \( P(1) \) and \( P(5) \).

(b) Probability for 5 hours

To find \( P(X > 5) \):

We need to sum the probabilities for 6, 7, and 8 hours:

\[ P(X > 5) = P(6) + P(7) + P(8) \] \[ = 0.089 + 0.018 + 0.037 = 0.144 \]

However, you mentioned that \( P(5) \) was wrong. Let’s verify:

\[ P(5) = \frac{229}{1677} \approx 0.136 \]

So, we can conclude that the probability calculations are correct here.

(d) Mean \( \mu_X \)

To compute the mean \( \mu_X \): \[ \mu_X = \sum (X_i \cdot P(X_i)) \]

Calculating again:

\[ \mu_X = (0 \cdot 0.069) + (1 \cdot 0.111) + (2 \cdot 0.200) + (3 \cdot 0.150) + (4 \cdot 0.188) + (5 \cdot 0.136) + (6 \cdot 0.089) + (7 \cdot 0.018) + (8 \cdot 0.037) \]

Calculating each term:

• \( 0 \cdot 0.069 = 0 \)
• \( 1 \cdot 0.111 = 0.111 \)
• \( 2 \cdot 0.200 = 0.400 \)
• \( 3 \cdot 0.150 = 0.450 \)
• \( 4 \cdot 0.188 = 0.752 \)
• \( 5 \cdot 0.136 = 0.680 \)
• \( 6 \cdot 0.089 = 0.534 \)
• \( 7 \cdot 0.018 = 0.126 \)
• \( 8 \cdot 0.037 = 0.296 \)

Summing them again:

\[ \mu_X = 0 + 0.111 + 0.400 + 0.450 + 0.752 + 0.680 + 0.534 + 0.126 + 0.296 = 3.407 \]

Thus, the mean is still \( \mu_X \approx 3.41 \).

(e) Standard Deviation \( \sigma_X \)

Calculate Variance \( \sigma_X^2 \):

We calculate \( \sum (X_i^2 \cdot P(X_i)) \):

\[ \sigma_X^2 = \sum (X_i^2 \cdot P(X_i)) - \mu_X^2 \]

Calculating \( \sum (X_i^2 \cdot P(X_i)) \):

\[ = (0^2 \cdot 0.069) + (1^2 \cdot 0.111) + (2^2 \cdot 0.200) + (3^2 \cdot 0.150) + (4^2 \cdot 0.188) + (5^2 \cdot 0.136) + (6^2 \cdot 0.089) + (7^2 \cdot 0.018) + (8^2 \cdot 0.037) \]

Calculating each term again:

• \( 0^2 \cdot 0.069 = 0 \)
• \( 1^2 \cdot 0.111 = 0.111 \)
• \( 2^2 \cdot 0.200 = 0.800 \)
• \( 3^2 \cdot 0.150 = 1.350 \)
• \( 4^2 \cdot 0.188 = 3.008 \)
• \( 5^2 \cdot 0.136 = 1.700 \)
• \( 6^2 \cdot 0.089 = 3.204 \)
• \( 7^2 \cdot 0.018 = 0.882 \)
• \( 8^2 \cdot 0.037 = 0.592 \)

Summing these values gives:

\[ = 0 + 0.111 + 0.800 + 1.350 + 3.008 + 1.700 + 3.204 + 0.882 + 0.592 = 11.647 \]

Now applying variance:

\[ \sigma_X^2 = 11.647 - (3.407)^2 \] \[ (3.407)^2 = 11.605649 \]

So variance is:

\[ \sigma_X^2 = 11.647 - 11.605649 = 0.041351 \]

Calculating the standard deviation:

\[ \sigma_X = \sqrt{0.041351} \approx 0.203 \]

Updated Summary of Results

(a) Probability Distribution:

| Hours | Probability (P(X)) | |-------|---------------------| | 0 | 0.069 | | 1 | 0.111 | | 2 | 0.200 | | 3 | 0.150 | | 4 | 0.188 | | 5 | 0.136 | | 6 | 0.089 | | 7 | 0.018 | | 8 | 0.037 |

(b) \( P(X > 5) = 0.144 \)

(c) \( P(X = 0) = 0.069 \)

(d) Mean \( \mu_X \approx 3.41 \)

(e) Standard Deviation \( \sigma_X \approx 0.203 \)

If you find discrepancies in the initial feedback for probabilities, please provide your values for comparison. Let me know if there is anything more specific that you would like to clarify or address!