Question

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.

Answer 1

Now find the probability that the standard is exceeded on more than one day

Answer 2

To find the probability that the standard is exceeded on at least one day, you need to calculate the complement of the event that the standard is not exceeded on any day.

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.

Answer 3

To find the probability that the standard is exceeded on more than one day, we need to consider the probabilities for \(X = 2\), \(X = 3\), and \(X = 4\) as these represent the days where the standard is exceeded more than once.

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.