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

x| 0 1 2 3 4
P(x)| 0.32 0.38 0.20 0.06 0.04

A) Find P(3)
B) Find P(2 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.
F) Compute the standard deviation OvX . Round the answer to three decimal places.

1 answer

To solve these problems, we will be working with the given probability distribution of \(X\) where \(X\) represents 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 provided as follows:

\[ \begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \ \hline P(x) & 0.32 & 0.38 & 0.20 & 0.06 & 0.04 \ \end{array} \]

A) Find \(P(3)\)

From the table, \( P(3) = 0.06 \).

B) Find \(P(2 \text{ or fewer})\)

To find \(P(2 \text{ or fewer})\), we sum the probabilities for \(x = 0, 1, 2\):

\[ P(2 \text{ or fewer}) = P(0) + P(1) + P(2) = 0.32 + 0.38 + 0.20 = 0.90 \]

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

To find \(P(X \geq 3)\), we can sum \(P(3)\) and \(P(4)\):

\[ P(X \geq 3) = P(3) + P(4) = 0.06 + 0.04 = 0.10 \]

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

To find \(P(X > 1)\), we can sum \(P(2), P(3),\) and \(P(4)\):

\[ P(X > 1) = P(2) + P(3) + P(4) = 0.20 + 0.06 + 0.04 = 0.30 \]

E) Compute the mean \(E(X)\).

The mean of a discrete random variable \(X\) is calculated as follows:

\[ E(X) = \sum (x \cdot P(x)) = 0 \cdot 0.32 + 1 \cdot 0.38 + 2 \cdot 0.20 + 3 \cdot 0.06 + 4 \cdot 0.04 \]

Calculating each term:

\[ E(X) = 0 + 0.38 + 0.40 + 0.18 + 0.16 = 1.12 \]

So, \(E(X) = 1.12\).

F) Compute the standard deviation \(O(X)\).

First, we need to calculate \(E(X^2)\):

\[ E(X^2) = \sum (x^2 \cdot P(x)) = 0^2 \cdot 0.32 + 1^2 \cdot 0.38 + 2^2 \cdot 0.20 + 3^2 \cdot 0.06 + 4^2 \cdot 0.04 \]

Calculating each term:

\[ E(X^2) = 0 + 0.38 + 0.80 + 0.54 + 0.64 = 2.36 \]

Now, we can calculate the variance \(Var(X)\):

\[ Var(X) = E(X^2) - (E(X))^2 = 2.36 - (1.12)^2 = 2.36 - 1.2544 = 1.1056 \]

Finally, the standard deviation \(O(X)\) is:

\[ O(X) = \sqrt{Var(X)} = \sqrt{1.1056} \approx 1.05 \]

Summary of Results:

  • \( P(3) = 0.06 \)
  • \( P(2 \text{ or fewer}) = 0.90 \)
  • \( P(X \geq 3) = 0.10 \)
  • \( P(X > 1) = 0.30 \)
  • \( E(X) = 1.12 \)
  • \( O(X) \approx 1.05 \)