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Sagot :
Sure, Isabella! Let's go through the solution step by step:
### Part (a): Verify that this is a discrete probability distribution
To verify that this is a discrete probability distribution, we need to check two conditions:
1. The sum of the probabilities must equal 1.
2. Each probability must be between 0 and 1, inclusive.
Let's check these conditions:
1. Sum of probabilities:
The probabilities given are:
[tex]\[ \begin{align*} P(X = 0) &= 0.1685 \\ P(X = 1) &= 0.3364 \\ P(X = 2) &= 0.2873 \\ P(X = 3) &= 0.1476 \\ P(X = 4) &= 0.0363 \\ P(X = 5) &= 0.0239 \\ \end{align*} \][/tex]
Now, let's sum these probabilities:
[tex]\[ 0.1685 + 0.3364 + 0.2873 + 0.1476 + 0.0363 + 0.0239 = 1.000 \][/tex]
The sum is exactly 1.
2. Check if each probability is between 0 and 1, inclusive:
[tex]\[ 0 \leq 0.1685 \leq 1 \\ 0 \leq 0.3364 \leq 1 \\ 0 \leq 0.2873 \leq 1 \\ 0 \leq 0.1476 \leq 1 \\ 0 \leq 0.0363 \leq 1 \\ 0 \leq 0.0239 \leq 1 \][/tex]
All given probabilities are between 0 and 1.
Since both conditions are satisfied, this is a discrete probability distribution.
### Part (b): Graph the probability distribution and describe its shape
To graph the probability distribution, we plot the probabilities [tex]\( P(X) \)[/tex] against the values of [tex]\( X \)[/tex].
Where:
- [tex]\( X \)[/tex] = {0, 1, 2, 3, 4, 5}
- Corresponding probabilities [tex]\( P(X) \)[/tex] = {0.1685, 0.3364, 0.2873, 0.1476, 0.0363, 0.0239}
We can create a bar graph where the x-axis represents the number of hits [tex]\( X \)[/tex] and the y-axis represents the probability [tex]\( P(X) \)[/tex].
Here is how it looks:
[tex]\[ \begin{array}{l|l} X & P(X) \\ \hline 0 & 0.1685 \\ 1 & 0.3364 \\ 2 & 0.2873 \\ 3 & 0.1476 \\ 4 & 0.0363 \\ 5 & 0.0239 \\ \end{array} \][/tex]
Description of the shape:
The shape of this probability distribution graph can be described as right-skewed (or positively skewed). This means that the bulk of the distribution is on the left, with a long tail to the right.
### Part (c) to (f) - Additional steps (not explicitly listed in the prompt but often included in full solutions)
Typically additional steps might include:
- (c) Calculating the mean (expected value) [tex]\( E(X) \)[/tex]
- (d) Calculating the variance [tex]\( \text{Var}(X) \)[/tex]
- (e) Calculating the standard deviation [tex]\( \sigma(X) \)[/tex]
But since we aren't asked for this, we can just leave it here.
To summarize, we've verified the distribution is correct and provided a description of its shape. Feel free to ask any more questions if needed!
### Part (a): Verify that this is a discrete probability distribution
To verify that this is a discrete probability distribution, we need to check two conditions:
1. The sum of the probabilities must equal 1.
2. Each probability must be between 0 and 1, inclusive.
Let's check these conditions:
1. Sum of probabilities:
The probabilities given are:
[tex]\[ \begin{align*} P(X = 0) &= 0.1685 \\ P(X = 1) &= 0.3364 \\ P(X = 2) &= 0.2873 \\ P(X = 3) &= 0.1476 \\ P(X = 4) &= 0.0363 \\ P(X = 5) &= 0.0239 \\ \end{align*} \][/tex]
Now, let's sum these probabilities:
[tex]\[ 0.1685 + 0.3364 + 0.2873 + 0.1476 + 0.0363 + 0.0239 = 1.000 \][/tex]
The sum is exactly 1.
2. Check if each probability is between 0 and 1, inclusive:
[tex]\[ 0 \leq 0.1685 \leq 1 \\ 0 \leq 0.3364 \leq 1 \\ 0 \leq 0.2873 \leq 1 \\ 0 \leq 0.1476 \leq 1 \\ 0 \leq 0.0363 \leq 1 \\ 0 \leq 0.0239 \leq 1 \][/tex]
All given probabilities are between 0 and 1.
Since both conditions are satisfied, this is a discrete probability distribution.
### Part (b): Graph the probability distribution and describe its shape
To graph the probability distribution, we plot the probabilities [tex]\( P(X) \)[/tex] against the values of [tex]\( X \)[/tex].
Where:
- [tex]\( X \)[/tex] = {0, 1, 2, 3, 4, 5}
- Corresponding probabilities [tex]\( P(X) \)[/tex] = {0.1685, 0.3364, 0.2873, 0.1476, 0.0363, 0.0239}
We can create a bar graph where the x-axis represents the number of hits [tex]\( X \)[/tex] and the y-axis represents the probability [tex]\( P(X) \)[/tex].
Here is how it looks:
[tex]\[ \begin{array}{l|l} X & P(X) \\ \hline 0 & 0.1685 \\ 1 & 0.3364 \\ 2 & 0.2873 \\ 3 & 0.1476 \\ 4 & 0.0363 \\ 5 & 0.0239 \\ \end{array} \][/tex]
Description of the shape:
The shape of this probability distribution graph can be described as right-skewed (or positively skewed). This means that the bulk of the distribution is on the left, with a long tail to the right.
### Part (c) to (f) - Additional steps (not explicitly listed in the prompt but often included in full solutions)
Typically additional steps might include:
- (c) Calculating the mean (expected value) [tex]\( E(X) \)[/tex]
- (d) Calculating the variance [tex]\( \text{Var}(X) \)[/tex]
- (e) Calculating the standard deviation [tex]\( \sigma(X) \)[/tex]
But since we aren't asked for this, we can just leave it here.
To summarize, we've verified the distribution is correct and provided a description of its shape. Feel free to ask any more questions if needed!
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