Answer :
To find the mean of a probability distribution, we use the formula for the expected value:
[tex]\[ \mu = \sum (x \cdot P(x)) \][/tex]
where [tex]\( x \)[/tex] represents each value of the random variable, and [tex]\( P(x) \)[/tex] represents the corresponding probability.
Given the values and their probabilities from the distribution table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 12 & 14 & 17 & 5 \\ \hline P(x) & 0.20 & 0.30 & 0.20 & 0.08 & 0.22 \\ \hline \end{array} \][/tex]
we calculate the mean as follows:
1. Multiply each value by its probability:
- [tex]\( 8 \times 0.20 = 1.60 \)[/tex]
- [tex]\( 12 \times 0.30 = 3.60 \)[/tex]
- [tex]\( 14 \times 0.20 = 2.80 \)[/tex]
- [tex]\( 17 \times 0.08 = 1.36 \)[/tex]
- [tex]\( 5 \times 0.22 = 1.10 \)[/tex]
2. Sum these products:
[tex]\[ 1.60 + 3.60 + 2.80 + 1.36 + 1.10 = 10.46 \][/tex]
Therefore, the mean of the distribution is:
[tex]\[ \mu = 10.46 \][/tex]
The correct answer is:
[tex]\[\text{10.46}\][/tex]
[tex]\[ \mu = \sum (x \cdot P(x)) \][/tex]
where [tex]\( x \)[/tex] represents each value of the random variable, and [tex]\( P(x) \)[/tex] represents the corresponding probability.
Given the values and their probabilities from the distribution table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 12 & 14 & 17 & 5 \\ \hline P(x) & 0.20 & 0.30 & 0.20 & 0.08 & 0.22 \\ \hline \end{array} \][/tex]
we calculate the mean as follows:
1. Multiply each value by its probability:
- [tex]\( 8 \times 0.20 = 1.60 \)[/tex]
- [tex]\( 12 \times 0.30 = 3.60 \)[/tex]
- [tex]\( 14 \times 0.20 = 2.80 \)[/tex]
- [tex]\( 17 \times 0.08 = 1.36 \)[/tex]
- [tex]\( 5 \times 0.22 = 1.10 \)[/tex]
2. Sum these products:
[tex]\[ 1.60 + 3.60 + 2.80 + 1.36 + 1.10 = 10.46 \][/tex]
Therefore, the mean of the distribution is:
[tex]\[ \mu = 10.46 \][/tex]
The correct answer is:
[tex]\[\text{10.46}\][/tex]