To find \( P(1) \) from the given probability distribution of \( X \), you need to identify the probability corresponding to \( X = 1 \). The probability distribution is given as:
\[
\begin{array}{c|ccccc}
X & 0 & 1 & 2 & 3 & 4 \\
\hline
P(X) & 0.31 & 0.38 & 0.21 & 0.07 & 0.03 \\
\end{array}
\]
From the table, the probability \( P(1) \) corresponds to the value when \( X = 1 \), which is 0.38.
So, \( P(1) = 0.38 \).
To find \( P(2 \text{ or fewer}) \), you need to sum the probabilities for \( X = 0 \), \( X = 1 \), and \( X = 2 \).
Given the probability distribution:
\[
\begin{array}{c|ccccc}
X & 0 & 1 & 2 & 3 & 4 \\
\hline
P(X) & 0.31 & 0.38 & 0.21 & 0.07 & 0.03 \\
\end{array}
\]
The probabilities are:
- \( P(0) = 0.31 \)
- \( P(1) = 0.38 \)
- \( P(2) = 0.21 \)
Now, sum these probabilities:
\[
P(2 \text{ or fewer}) = P(0) + P(1) + P(2) = 0.31 + 0.38 + 0.21 = 0.90
\]
So, \( P(2 \text{ or fewer}) = 0.90 \).
Knowing all of this. Find the probability that the standard is exceeded on at least one day.
3 answers
The probability that the standard is not exceeded on any day is when \(X = 0\), so \(P(\text{standard not exceeded on any day}) = P(X = 0) = 0.31\).
Therefore, the probability that the standard is exceeded on at least one day is given by:
\[P(\text{standard exceeded on at least one day}) = 1 - P(\text{standard not exceeded on any day}) = 1 - P(X = 0) = 1 - 0.31 = 0.69\]
So, the probability that the standard is exceeded on at least one day is 0.69.
Given the probability distribution:
\[
\begin{array}{c|ccccc}
X & 0 & 1 & 2 & 3 & 4 \\
\hline
P(X) & 0.31 & 0.38 & 0.21 & 0.07 & 0.03 \\
\end{array}
\]
The probabilities we are interested in are:
- \(P(2) = 0.21\)
- \(P(3) = 0.07\)
- \(P(4) = 0.03\)
To find the probability that the standard is exceeded on more than one day, we sum these probabilities:
\[P(\text{standard exceeded on more than one day}) = P(X = 2) + P(X = 3) + P(X = 4) = 0.21 + 0.07 + 0.03 = 0.31\]
Therefore, the probability that the standard is exceeded on more than one day is 0.31.