A recent survey asked 824 people how many days they would wait to seek medical treatment if they were suffering pain that interfered with their ability to work. The results are presented in the following table.

Question

Number of days
0 with a frequency of 29
1 with a frequency of 435
2 with a frequency of 265
3 with a frequency of 70
4 with a frequency of 15
5 with a frequency of 10

Consider these 824 people to be a population. Let X be the number of days for a person sampled at random from this population.
(a) Construct the probability distribution of X. Round the answer to three decimal places.
P| 0 1 2 3 4 5
P(x)| [?] [?] [?] [?] [?]

Answer 1

To construct the probability distribution of \( X \), we need to calculate the probabilities for each number of days \( (0, 1, 2, 3, 4, 5) \) based on the given frequencies.

The probability \( P(X = k) \) for each \( k \) is given by the formula:

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

Given the total number of respondents is 824, we can calculate \( P(X) \) for each day as follows:

1. For \( X = 0 \): \[ P(X = 0) = \frac{29}{824} \approx 0.035 \]

2. For \( X = 1 \): \[ P(X = 1) = \frac{435}{824} \approx 0.528 \]

3. For \( X = 2 \): \[ P(X = 2) = \frac{265}{824} \approx 0.321 \]

4. For \( X = 3 \): \[ P(X = 3) = \frac{70}{824} \approx 0.085 \]

5. For \( X = 4 \): \[ P(X = 4) = \frac{15}{824} \approx 0.018 \]

6. For \( X = 5 \): \[ P(X = 5) = \frac{10}{824} \approx 0.012 \]

Now we can summarize the probability distribution of \( X \):

\[ \begin{aligned} P(X = 0) & \approx 0.035 \ P(X = 1) & \approx 0.528 \ P(X = 2) & \approx 0.321 \ P(X = 3) & \approx 0.085 \ P(X = 4) & \approx 0.018 \ P(X = 5) & \approx 0.012 \ \end{aligned} \]

In table format, this can be presented as:

\[ \begin{array}{c|c} X & P(X) \ \hline 0 & 0.035 \ 1 & 0.528 \ 2 & 0.321 \ 3 & 0.085 \ 4 & 0.018 \ 5 & 0.012 \ \end{array} \]

Thus, the final probability distribution rounded to three decimal places is:

\[ \begin{aligned} P(0) & = 0.035 \ P(1) & = 0.528 \ P(2) & = 0.321 \ P(3) & = 0.085 \ P(4) & = 0.018 \ P(5) & = 0.012 \ \end{aligned} \]

Answer 2

b) Find the probability that a person would wait for 2 days.

Answer 3

To find the probability that a person would wait for 2 days, we refer to the probability distribution we constructed in part (a).

From the probability distribution:

\[ P(X = 2) \approx 0.321 \]

Thus, the probability that a person would wait for 2 days is:

\[ \boxed{0.321} \]

Answer 4

(c) Find the probability that a person would wait for more than 2 days.

Answer 5

To find the probability that a person would wait for more than 2 days, we need to sum the probabilities for \( X = 3, 4, \) and \( 5 \).

From the probability distribution we previously calculated:

• \( P(X = 3) \approx 0.085 \)
• \( P(X = 4) \approx 0.018 \)
• \( P(X = 5) \approx 0.012 \)

Now, we can calculate the total probability for waiting more than 2 days:

\[ P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) \]

Substituting the values:

\[ P(X > 2) \approx 0.085 + 0.018 + 0.012 \approx 0.115 \]

Thus, the probability that a person would wait for more than 2 days is:

\[ \boxed{0.115} \]

Answer 6

(d) Compute the mean UvX . Round the answer to two decimal places.
e) Compute the standard deviation OvX. Round the answer to 3 decimal places.

Answer 7

To calculate the mean \( \mu_X \) and the standard deviation \( \sigma_X \) of the probability distribution, we first need to use the formulas for the mean and standard deviation.

(d) Mean \( \mu_X \)

The mean \( \mu_X \) is calculated using the formula:

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

Using the probabilities we calculated earlier:

\[ \begin{align*} \mu_X & = (0 \cdot 0.035) + (1 \cdot 0.528) + (2 \cdot 0.321) + (3 \cdot 0.085) + (4 \cdot 0.018) + (5 \cdot 0.012) \ & = 0 + 0.528 + 0.642 + 0.255 + 0.072 + 0.060 \ & = 1.557 \end{align*} \]

So, rounding to two decimal places, the mean \( \mu_X \) is:

\[ \boxed{1.56} \]

(e) Standard Deviation \( \sigma_X \)

The standard deviation is calculated using the formula:

\[ \sigma_X = \sqrt{\sum ((x - \mu_X)^2 \cdot P(X = x))} \]

First, we can compute \( (x - \mu_X)^2 \) for each \( x \):

\[ \begin{align*} (0 - 1.557)^2 & \approx 2.426 \ (1 - 1.557)^2 & \approx 0.313 \ (2 - 1.557)^2 & \approx 0.195 \ (3 - 1.557)^2 & \approx 2.054 \ (4 - 1.557)^2 & \approx 5.970 \ (5 - 1.557)^2 & \approx 11.849 \ \end{align*} \]

Now, we calculate:

\[ \begin{align*} \sigma_X^2 & = (0.035 \cdot 2.426) + (0.528 \cdot 0.313) + (0.321 \cdot 0.195) + (0.085 \cdot 2.054) + (0.018 \cdot 5.970) + (0.012 \cdot 11.849) \ & = 0.085 && \text{(from } P(X=0) \text{)} \ & + 0.165 && \text{(from } P(X=1) \text{)} \ & + 0.062 && \text{(from } P(X=2) \text{)} \ & + 0.175 && \text{(from } P(X=3) \text{)} \ & + 0.107 && \text{(from } P(X=4) \text{)} \ & + 0.142 && \text{(from } P(X=5) \text{)} \ & = 0.716 \end{align*} \]

Now take the square root to find \( \sigma_X \):

\[ \sigma_X = \sqrt{0.716} \approx 0.846 \]

Thus, rounding to three decimal places, the standard deviation \( \sigma_X \) is:

\[ \boxed{0.846} \]