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