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The federal government has enacted maximum allowable standards for air pollutants such as ozone. Let X be the number of days per year that the level of air pollution exceeds the standard in a certain city. The probability distribution of X is given by

Find P(1)
Find P(3 or fewer)
(c) Find the probability that the standard is exceeded on at least three days.
(d) Find the probability that the standard is exceeded on more than one day.
(e) Compute the mean UvX. Round the answer to two decimal places.

7 answers

To answer your questions regarding the random variable X, which represents the number of days per year that the level of air pollution exceeds the federal standards, we need the specific probability distribution for X (such as a table or a formula).

Since you have not provided the probability distribution, I will provide a general approach you can take depending on what type of distribution it is (e.g., Binomial, Poisson):

  1. Find P(1) - This requires finding the probability that X = 1. Using the probability mass function (PMF) from your distribution, look for the value corresponding to X = 1.

  2. Find P(3 or fewer) - This requires calculating the cumulative probability for X values 0, 1, 2, and 3: \[ P(X \leq 3) = P(0) + P(1) + P(2) + P(3) \]

  3. Find the probability that the standard is exceeded on at least three days: \[ P(X \geq 3) = 1 - P(X \leq 2) \] or directly calculate \( P(3) + P(4) + \ldots \).

  4. Find the probability that the standard is exceeded on more than one day: \[ P(X > 1) = 1 - P(X \leq 1) = 1 - [P(0) + P(1)] \]

  5. Compute the mean \( \mu_X \) - The mean or expected value of a discrete random variable can be computed using: \[ \mu_X = \sum_{x=0}^{n} x \cdot P(x) \] Here, "n" is the maximum value of X and the sums are computed for each value of X with their respective probabilities.

To get specific values for each of the above steps, you will need to provide the probability distribution of X. Please share that information to enable detailed computations!

X| 0 1 2 3 4
P(x)| .33 .39 .18 .06 .04

Thank you for providing the probability distribution for \( X \), the number of days per year that the level of air pollution exceeds the standard. The provided probabilities are as follows:

\[ \begin{array}{|c|c|} \hline X & P(X) \ \hline 0 & 0.33 \ 1 & 0.39 \ 2 & 0.18 \ 3 & 0.06 \ 4 & 0.04 \ \hline \end{array} \]

Let's calculate the required probabilities and the mean.

a) Find \( P(1) \)

From the table, we can directly read: \[ P(1) = 0.39 \]

b) Find \( P(3 \text{ or fewer}) \)

This is calculated as \[ P(X \leq 3) = P(0) + P(1) + P(2) + P(3) \] Calculating it: \[ P(X \leq 3) = 0.33 + 0.39 + 0.18 + 0.06 = 0.96 \]

c) Find the probability that the standard is exceeded on at least three days

This is given by \[ P(X \geq 3) = P(3) + P(4) \] Calculating it: \[ P(X \geq 3) = 0.06 + 0.04 = 0.10 \]

d) Find the probability that the standard is exceeded on more than one day

This can be calculated as \[ P(X > 1) = 1 - P(X \leq 1) \] Thus, \[ P(X \leq 1) = P(0) + P(1) = 0.33 + 0.39 = 0.72 \] Therefore: \[ P(X > 1) = 1 - 0.72 = 0.28 \]

e) Compute the mean \( \mu_X \)

The mean \( \mu_X \) is calculated using the formula: \[ \mu_X = \sum_{x=0}^{n} x \cdot P(x) \] Calculating it step-by-step: \[ \mu_X = (0 \cdot 0.33) + (1 \cdot 0.39) + (2 \cdot 0.18) + (3 \cdot 0.06) + (4 \cdot 0.04) \] Breaking it down: \[ = 0 + 0.39 + 0.36 + 0.18 + 0.16 \] Now sum these: \[ = 0.39 + 0.36 + 0.18 + 0.16 = 1.09 \]

Therefore, the answers are:

  • \( P(1) = 0.39 \)
  • \( P(3 \text{ or fewer}) = 0.96 \)
  • \( P(X \geq 3) = 0.10 \)
  • \( P(X > 1) = 0.28 \)
  • Mean \( \mu_X = 1.09 \)

Please confirm whether you need any further assistance!

(f) Compute the standard deviation OvX . Round the answer to three decimal places.

To compute the standard deviation \( \sigma_X \) of the random variable \( X \), we first need to find the variance \( \sigma^2_X \). The variance can be calculated using the formula:

\[ \sigma^2_X = E(X^2) - (E(X))^2 \]

Where:

  • \( E(X) \) is the expected value (which we previously calculated as \( \mu_X = 1.09 \)).
  • \( E(X^2) \) is the expected value of \( X^2 \).

Step 1: Calculate \( E(X^2) \)

To find \( E(X^2) \), we use the formula:

\[ E(X^2) = \sum_{x=0}^{n} x^2 \cdot P(x) \]

Using the provided probabilities: \[ E(X^2) = (0^2 \cdot 0.33) + (1^2 \cdot 0.39) + (2^2 \cdot 0.18) + (3^2 \cdot 0.06) + (4^2 \cdot 0.04) \]

Calculating each term:

  • \( 0^2 \cdot 0.33 = 0 \)
  • \( 1^2 \cdot 0.39 = 0.39 \)
  • \( 2^2 \cdot 0.18 = 4 \cdot 0.18 = 0.72 \)
  • \( 3^2 \cdot 0.06 = 9 \cdot 0.06 = 0.54 \)
  • \( 4^2 \cdot 0.04 = 16 \cdot 0.04 = 0.64 \)

Adding these up: \[ E(X^2) = 0 + 0.39 + 0.72 + 0.54 + 0.64 = 1.29 \]

Step 2: Calculate the Variance \( \sigma^2_X \)

Now we can calculate the variance: \[ \sigma^2_X = E(X^2) - (E(X))^2 \] Substituting in our values: \[ \sigma^2_X = 1.29 - (1.09)^2 \] Calculating \( (1.09)^2 \): \[ (1.09)^2 = 1.1881 \] Now substitute: \[ \sigma^2_X = 1.29 - 1.1881 = 0.1019 \]

Step 3: Calculate the Standard Deviation \( \sigma_X \)

Finally, take the square root of the variance to find the standard deviation: \[ \sigma_X = \sqrt{\sigma^2_X} = \sqrt{0.1019} \approx 0.319 \]

Final Answer:

The standard deviation \( \sigma_X \) rounded to three decimal places is:

\[ \sigma_X \approx 0.319 \]

If you have further questions or need assistance with anything else, feel free to ask!

The answer is not 0.319

Let's go through the calculations for the variance and standard deviation again step by step.

Step 1: Calculate \(E(X)\) Again

We already found: \[ E(X) = \mu_X = 1.09 \]

Step 2: Calculate \(E(X^2)\)

We compute \(E(X^2)\) again using:

\[ E(X^2) = \sum_{x=0}^{n} x^2 \cdot P(x) \]

Using the probabilities provided:

  • \(P(0) = 0.33\)
  • \(P(1) = 0.39\)
  • \(P(2) = 0.18\)
  • \(P(3) = 0.06\)
  • \(P(4) = 0.04\)

Now we calculate: \[ E(X^2) = (0^2 \cdot 0.33) + (1^2 \cdot 0.39) + (2^2 \cdot 0.18) + (3^2 \cdot 0.06) + (4^2 \cdot 0.04) \]

Calculating each term:

  • \(0^2 \cdot 0.33 = 0\)
  • \(1^2 \cdot 0.39 = 0.39\)
  • \(2^2 \cdot 0.18 = 4 \cdot 0.18 = 0.72\)
  • \(3^2 \cdot 0.06 = 9 \cdot 0.06 = 0.54\)
  • \(4^2 \cdot 0.04 = 16 \cdot 0.04 = 0.64\)

Adding these: \[ E(X^2) = 0 + 0.39 + 0.72 + 0.54 + 0.64 = 1.29 \]

Step 3: Calculate Variance \( \sigma^2_X \)

Using the formula: \[ \sigma^2_X = E(X^2) - (E(X))^2 \] Substituting our values: \[ \sigma^2_X = 1.29 - (1.09)^2 \] Calculating \( (1.09)^2 \): \[ (1.09)^2 = 1.1881 \]

Now substituting back: \[ \sigma^2_X = 1.29 - 1.1881 = 0.1019 \]

Step 4: Calculate Standard Deviation \( \sigma_X \)

Taking the square root of the variance: \[ \sigma_X = \sqrt{0.1019} \approx 0.319 \]

Reevaluation

Given your statement that this is not the correct answer, let’s check the calculations again with precise values.

  1. Confirm calculations:
    • \(E(X^2) = 1.29\)
    • \(E(X) \approx 1.09\)
    • \((E(X))^2 \approx 1.1881\)

So: \[ \sigma^2_X = 1.29 - 1.1881 \] Calculating: \[ \sigma^2_X = 1.29 - 1.1881 = 0.1019 \] Calculating: \[ \sigma_X = \sqrt{0.1019} \approx 0.319 \]

Check Again with Precise Numbers

To ensure there are no calculation errors:

  1. Using a calculator for the square root: \[ \sigma_X = \sqrt{0.1019} = 0.319 \]

If you would like to provide additional context or if there is a specified value you know for the variance or standard deviation, please share! Otherwise, based on the calculations, the standard deviation should indeed be about 0.319.