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A recent survey asked 1674 people how many hours per day they were able to relax. The results are presented in the following table

Number of Hours 0, 1, 2, 3, 4, 5, 6, 7, 8
Frequency 115, 184, 338, 254, 315, 230, 148, 32, 58

Consider these 1674 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 6 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.

5 answers

To construct the probability distribution and calculate the requested probabilities and statistics, we will first summarize the data provided.

(a) Construct the probability distribution of X

To create a probability distribution, we will find the probability associated with each number of hours of relaxation (X):

\[ P(X = x) = \frac{\text{Frequency of } x}{\text{Total number of people}} \]

The total number of people is 1674.

Here is how we calculate it for each value of X:

| Number of Hours (X) | Frequency | Probability \( P(X=x) \) | |---------------------|-----------|---------------------------| | 0 | 115 | \( P(X=0) = \frac{115}{1674} \approx 0.069 \) | | 1 | 184 | \( P(X=1) = \frac{184}{1674} \approx 0.110 \) | | 2 | 338 | \( P(X=2) = \frac{338}{1674} \approx 0.202 \) | | 3 | 254 | \( P(X=3) = \frac{254}{1674} \approx 0.152 \) | | 4 | 315 | \( P(X=4) = \frac{315}{1674} \approx 0.188 \) | | 5 | 230 | \( P(X=5) = \frac{230}{1674} \approx 0.137 \) | | 6 | 148 | \( P(X=6) = \frac{148}{1674} \approx 0.088 \) | | 7 | 32 | \( P(X=7) = \frac{32}{1674} \approx 0.019 \) | | 8 | 58 | \( P(X=8) = \frac{58}{1674} \approx 0.035 \) |

Rounding these to three decimal places gives us the probability distribution:

| Number of Hours (X) | Probability \( P(X=x) \) | |---------------------|---------------------------| | 0 | 0.069 | | 1 | 0.110 | | 2 | 0.202 | | 3 | 0.152 | | 4 | 0.188 | | 5 | 0.137 | | 6 | 0.088 | | 7 | 0.019 | | 8 | 0.035 |

(b) Find the probability that a person relaxes more than 6 hours per day.

To find this probability, we will add the probabilities for X = 7 and X = 8:

\[ P(X > 6) = P(X = 7) + P(X = 8) \] \[ P(X > 6) = 0.019 + 0.035 = 0.054 \]

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

This is simply:

\[ P(X = 0) = 0.069 \]

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

To calculate the mean \( \mu \), we use the formula:

\[ \mu = \sum (x \cdot P(X=x)) \]

Calculating this:

\[ \mu = (0 \cdot 0.069) + (1 \cdot 0.110) + (2 \cdot 0.202) + (3 \cdot 0.152) + (4 \cdot 0.188) + (5 \cdot 0.137) + (6 \cdot 0.088) + (7 \cdot 0.019) + (8 \cdot 0.035) \] \[ = 0 + 0.110 + 0.404 + 0.456 + 0.752 + 0.685 + 0.528 + 0.133 + 0.280 \] \[ = 3.538 \]

Rounding to two decimal places gives us:

\[ \mu \approx 3.54 \]

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

First, we find the variance \( \sigma^2 \) using the formula:

\[ \sigma^2 = \sum \left( (x - \mu)^2 \cdot P(X=x) \right) \]

Calculating this:

\[ = (0 - 3.538)^2 \cdot 0.069 + (1 - 3.538)^2 \cdot 0.110 + (2 - 3.538)^2 \cdot 0.202 + (3 - 3.538)^2 \cdot 0.152 + (4 - 3.538)^2 \cdot 0.188 + (5 - 3.538)^2 \cdot 0.137 + (6 - 3.538)^2 \cdot 0.088 + (7 - 3.538)^2 \cdot 0.019 + (8 - 3.538)^2 \cdot 0.035 \]

Calculating each term:

  1. \( (0 - 3.538)^2 \cdot 0.069 \approx 0.037538 \)
  2. \( (1 - 3.538)^2 \cdot 0.110 \approx 0.050752 \)
  3. \( (2 - 3.538)^2 \cdot 0.202 \approx 0.028831 \)
  4. \( (3 - 3.538)^2 \cdot 0.152 \approx 0.012477 \)
  5. \( (4 - 3.538)^2 \cdot 0.188 \approx 0.020677 \)
  6. \( (5 - 3.538)^2 \cdot 0.137 \approx 0.056162 \)
  7. \( (6 - 3.538)^2 \cdot 0.088 \approx 0.059583 \)
  8. \( (7 - 3.538)^2 \cdot 0.019 \approx 0.052571 \)
  9. \( (8 - 3.538)^2 \cdot 0.035 \approx 0.078811 \)

Adding these up:

\[ \sigma^2 = 0.037538 + 0.050752 + 0.028831 + 0.012477 + 0.020677 + 0.056162 + 0.059583 + 0.052571 + 0.078811 \approx 0.396 \]

To find the standard deviation, take the square root:

\[ \sigma \approx \sqrt{0.396} \approx 0.629 \]

Rounding to three decimal places gives us:

\[ \sigma \approx 0.629 \]

Summary of Results

(a) Probability Distribution:

  • \( P(X=0) = 0.069 \)
  • \( P(X=1) = 0.110 \)
  • \( P(X=2) = 0.202 \)
  • \( P(X=3) = 0.152 \)
  • \( P(X=4) = 0.188 \)
  • \( P(X=5) = 0.137 \)
  • \( P(X=6) = 0.088 \)
  • \( P(X=7) = 0.019 \)
  • \( P(X=8) = 0.035 \)

(b) \( P(X > 6) \approx 0.054 \)

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

(d) Mean \( UvX \approx 3.54 \)

(e) Standard Deviation \( OvX \approx 0.629 \)

D is 3.35
Knowing this, now compute the standard deviation OvX

Given that the mean \( \mu \) (or \( UvX \)) is now stated to be \( 3.35 \), we can recalculate the standard deviation \( \sigma \) (or \( OvX \)) using the updated mean.

Standard Deviation Calculation

The formula for the variance \( \sigma^2 \) is:

\[ \sigma^2 = \sum \left( (x - \mu)^2 \cdot P(X=x) \right) \]

We will use the same probabilities from the distribution table and adjust the calculation based on the new mean \( \mu = 3.35 \).

  1. Calculate each term \( (x - \mu)^2 \cdot P(X=x) \):

\[ \begin{align*} P(X=0) & : (0 - 3.35)^2 \cdot 0.069 \ & = 11.2225 \cdot 0.069 \approx 0.7734 \ \ P(X=1) & : (1 - 3.35)^2 \cdot 0.110 \ & = 5.6225 \cdot 0.110 \approx 0.6185 \ \ P(X=2) & : (2 - 3.35)^2 \cdot 0.202 \ & = 1.8225 \cdot 0.202 \approx 0.3685 \ \ P(X=3) & : (3 - 3.35)^2 \cdot 0.152 \ & = 0.1225 \cdot 0.152 \approx 0.0186 \ \ P(X=4) & : (4 - 3.35)^2 \cdot 0.188 \ & = 0.4225 \cdot 0.188 \approx 0.0795 \ \ P(X=5) & : (5 - 3.35)^2 \cdot 0.137 \ & = 2.7225 \cdot 0.137 \approx 0.3726 \ \ P(X=6) & : (6 - 3.35)^2 \cdot 0.088 \ & = 7.1125 \cdot 0.088 \approx 0.6240 \ \ P(X=7) & : (7 - 3.35)^2 \cdot 0.019 \ & = 13.3225 \cdot 0.019 \approx 0.2531 \ \ P(X=8) & : (8 - 3.35)^2 \cdot 0.035 \ & = 21.6025 \cdot 0.035 \approx 0.7561 \ \end{align*} \]

  1. Add all of these values to find the variance \( \sigma^2 \):

\[ \sigma^2 = 0.7734 + 0.6185 + 0.3685 + 0.0186 + 0.0795 + 0.3726 + 0.6240 + 0.2531 + 0.7561 \]

Calculating the sum:

\[ \sigma^2 \approx 3.4943 \]

  1. Finally, compute the standard deviation \( \sigma \):

\[ \sigma = \sqrt{3.4943} \approx 1.870 \]

Conclusion

The standard deviation \( OvX \) is approximately 1.870 when rounded to three decimal places.

No, E is 1.870

Thank you for your patience. Given that the standard deviation \( OvX \) is now stated to be \( 1.870 \), let’s clarify the calculations to ensure we derive the proper standard deviation based on the updated mean \( \mu = 3.35 \) from our previous analysis.

We'll recalculate the standard deviation step by step using the correct \( \mu \) and verify the calculations again.

Standard Deviation Formula

The variance \( \sigma^2 \) is defined as:

\[ \sigma^2 = \sum_{i} ((x_i - \mu)^2 \cdot P(X = x_i)) \]

Where \( x_i \) are the values of \( X \), \( \mu \) is the mean, and \( P(X = x_i) \) are their respective probabilities.

Probability Distribution Recap

Using the previously calculated probabilities:

| Number of Hours (X) | Probability \( P(X=x) \) | |---------------------|---------------------------| | 0 | 0.069 | | 1 | 0.110 | | 2 | 0.202 | | 3 | 0.152 | | 4 | 0.188 | | 5 | 0.137 | | 6 | 0.088 | | 7 | 0.019 | | 8 | 0.035 |

Recalculating Variance

Now let's recalculate \( \sigma^2 \):

  1. Calculate each \( (x_i - \mu)^2 \cdot P(X=x_i) \):

\[ \begin{align*} P(X=0) & : (0 - 3.35)^2 \cdot 0.069 = 11.2225 \cdot 0.069 \approx 0.7744 \ \ P(X=1) & : (1 - 3.35)^2 \cdot 0.110 = 5.6225 \cdot 0.110 \approx 0.6185 \ \ P(X=2) & : (2 - 3.35)^2 \cdot 0.202 = 1.8225 \cdot 0.202 \approx 0.3684 \ \ P(X=3) & : (3 - 3.35)^2 \cdot 0.152 = 0.1225 \cdot 0.152 \approx 0.0186 \ \ P(X=4) & : (4 - 3.35)^2 \cdot 0.188 = 0.4225 \cdot 0.188 \approx 0.0795 \ \ P(X=5) & : (5 - 3.35)^2 \cdot 0.137 = 2.7225 \cdot 0.137 \approx 0.3725 \ \ P(X=6) & : (6 - 3.35)^2 \cdot 0.088 = 7.1125 \cdot 0.088 \approx 0.6240 \ \ P(X=7) & : (7 - 3.35)^2 \cdot 0.019 = 13.3225 \cdot 0.019 \approx 0.2531 \ \ P(X=8) & : (8 - 3.35)^2 \cdot 0.035 = 21.6025 \cdot 0.035 \approx 0.7561 \ \end{align*} \]

  1. Combine these results to find the variance:

\[ \begin{align*} \sigma^2 & = 0.7744 + 0.6185 + 0.3684 + 0.0186 + 0.0795 + 0.3725 + 0.6240 + 0.2531 + 0.7561 \ & \approx 3.4946 \end{align*} \]

  1. Taking the square root gives the standard deviation:

\[ \sigma = \sqrt{3.4946} \approx 1.870 \]

Conclusion

The standard deviation \( OvX \) is indeed approximately 1.870 when rounded to three decimal places, confirming your input. Apologies for any prior confusion, and let me know if there are any more calculations or clarifications needed!