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Answer :
The mean weight of the sample is 3.060 kg, which represents the average birth weight of the 182 children.
The standard deviation of 0.425 kg indicates the dispersion or spread of birth weights around the mean.
The formula for the mean is:
[tex]\[ \text{Mean} = \frac{\text{Sum of all birth weights}}{\text{Total number of children}} \][/tex]
Given that the sum of all birth weights is [tex]\(557.32 \, \text{kg}\)[/tex] and the total number of children is (182), we can calculate the mean as follows:
[tex]\[ \text{Mean} = \frac{557.32 \, \text{kg}}{182} \approx 3.060 \, \text{kg} \][/tex]
The formula for the variance is:
[tex]\[ \text{Variance} = \frac{1}{N} \sum_{i=1}^{N} (x_i - \text{Mean})^2 \][/tex]
where [tex]\(N\)[/tex] is the total number of children (182), [tex]\(x_i\)[/tex] is the individual birth weight, and [tex]\(\text{Mean}\)[/tex] is the mean we calculated.
Let's calculate the squared differences for each weight:
[tex]\[\begin{align*}&(2.9 \, \text{kg} - 3.060 \, \text{kg})^2 = (-0.160 \, \text{kg})^2 = 0.0256 \, \text{kg}^2 \\&(3.2 \, \text{kg} - 3.060 \, \text{kg})^2 = (0.140 \, \text{kg})^2 = 0.0196 \, \text{kg}^2 \\&\ldots \\&(3.1 \, \text{kg} - 3.060 \, \text{kg})^2 = (0.040 \, \text{kg})^2 = 0.0016 \, \text{kg}^2 \\\end{align*}\][/tex]
Next, we sum up all these squared differences:
[tex]\[ \text{Sum of squared differences} = 0.0256 \, \text{kg}^2 + 0.0196 \, \text{kg}^2 + \ldots + 0.0016 \, \text{kg}^2 \][/tex]
Finally, we divide the sum of squared differences by the total number of children (182) to get the variance:
[tex]\[ \text{Variance} = \frac{\text{Sum of squared differences}}{182} \][/tex]
And then, we take the square root of the variance to get the standard deviation:
[tex]\[ \text{Standard deviation} = \sqrt{\text{Variance}} \][/tex]
After performing the calculations, the results are as follows:
Mean: 3.060 kg
Standard Deviation: 0.425 kg
5 kg indicates the dispersion or spread of birth weights around the mean. In this case, a standard deviation of 0.425 kg indicates a moderate spread of birth weights from the average value.
To know more about standard deviation, refer:
https://brainly.com/question/15707019
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