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Sagot :
To compare the centers of the distributions represented by [tex]\( H \)[/tex] and [tex]\( C \)[/tex], we need to calculate their expected values. The expected value (or mean) of a discrete random variable is calculated by summing the products of each value the variable can take and the probability of that value.
### Calculating the Expected Number of Eggs for Hannah's Hens
The probability distribution for the number of eggs laid by Hannah's hens [tex]\( P(H) \)[/tex] is given as:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Number of Eggs} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Probability} & 0.02 & 0.03 & 0.07 & 0.12 & 0.30 & 0.28 & 0.18 \\ \hline \end{array} \][/tex]
To find the expected value [tex]\( E(H) \)[/tex], we calculate:
[tex]\[ E(H) = \sum ( \text{Number of Eggs} \times \text{Probability}) \][/tex]
[tex]\[ E(H) = 0 \times 0.02 + 1 \times 0.03 + 2 \times 0.07 + 3 \times 0.12 + 4 \times 0.30 + 5 \times 0.28 + 6 \times 0.18 \][/tex]
[tex]\[ E(H) = 0 + 0.03 + 0.14 + 0.36 + 1.20 + 1.40 + 1.08 = 4.21 \][/tex]
Thus, the expected number of eggs laid by Hannah's hens in a day is 4.21.
### Calculating the Expected Number of Eggs for Claire's Hens
The probability distribution for the number of eggs laid by Claire's hens [tex]\( P(C) \)[/tex] is given as:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Number of Eggs} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Probability} & 0.05 & 0.09 & 0.14 & 0.23 & 0.21 & 0.16 & 0.12 \\ \hline \end{array} \][/tex]
To find the expected value [tex]\( E(C) \)[/tex], we calculate:
[tex]\[ E(C) = \sum ( \text{Number of Eggs} \times \text{Probability}) \][/tex]
[tex]\[ E(C) = 0 \times 0.05 + 1 \times 0.09 + 2 \times 0.14 + 3 \times 0.23 + 4 \times 0.21 + 5 \times 0.16 + 6 \times 0.12 \][/tex]
[tex]\[ E(C) = 0 + 0.09 + 0.28 + 0.69 + 0.84 + 0.80 + 0.72 = 3.42 \][/tex]
Thus, the expected number of eggs laid by Claire's hens in a day is 3.42.
### Conclusion
Based on the calculated expected values:
- Hannah's hens are expected to lay, on average, 4.21 eggs per day.
- Claire's hens are expected to lay, on average, 3.42 eggs per day.
Thus, the correct statement is:
- "Hannah's hens appear to lay more eggs, on average, than Claire's hens."
### Calculating the Expected Number of Eggs for Hannah's Hens
The probability distribution for the number of eggs laid by Hannah's hens [tex]\( P(H) \)[/tex] is given as:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Number of Eggs} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Probability} & 0.02 & 0.03 & 0.07 & 0.12 & 0.30 & 0.28 & 0.18 \\ \hline \end{array} \][/tex]
To find the expected value [tex]\( E(H) \)[/tex], we calculate:
[tex]\[ E(H) = \sum ( \text{Number of Eggs} \times \text{Probability}) \][/tex]
[tex]\[ E(H) = 0 \times 0.02 + 1 \times 0.03 + 2 \times 0.07 + 3 \times 0.12 + 4 \times 0.30 + 5 \times 0.28 + 6 \times 0.18 \][/tex]
[tex]\[ E(H) = 0 + 0.03 + 0.14 + 0.36 + 1.20 + 1.40 + 1.08 = 4.21 \][/tex]
Thus, the expected number of eggs laid by Hannah's hens in a day is 4.21.
### Calculating the Expected Number of Eggs for Claire's Hens
The probability distribution for the number of eggs laid by Claire's hens [tex]\( P(C) \)[/tex] is given as:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Number of Eggs} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Probability} & 0.05 & 0.09 & 0.14 & 0.23 & 0.21 & 0.16 & 0.12 \\ \hline \end{array} \][/tex]
To find the expected value [tex]\( E(C) \)[/tex], we calculate:
[tex]\[ E(C) = \sum ( \text{Number of Eggs} \times \text{Probability}) \][/tex]
[tex]\[ E(C) = 0 \times 0.05 + 1 \times 0.09 + 2 \times 0.14 + 3 \times 0.23 + 4 \times 0.21 + 5 \times 0.16 + 6 \times 0.12 \][/tex]
[tex]\[ E(C) = 0 + 0.09 + 0.28 + 0.69 + 0.84 + 0.80 + 0.72 = 3.42 \][/tex]
Thus, the expected number of eggs laid by Claire's hens in a day is 3.42.
### Conclusion
Based on the calculated expected values:
- Hannah's hens are expected to lay, on average, 4.21 eggs per day.
- Claire's hens are expected to lay, on average, 3.42 eggs per day.
Thus, the correct statement is:
- "Hannah's hens appear to lay more eggs, on average, than Claire's hens."
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