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The weights of ten young turkeys to the nearest 0.1 kg are: 0.8, 1.1, 1.2, 0.9, 1.2, 1.2, 0.9, 0.7, 1.0, 1.1 a. Find the mean and standard deviation for the turkeys. b. After being fed a special diet for one month, the weights of the turkeys doubled. Find the new mean and standard deviation. c. Comment, in general terms, on your findings from a and b.

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(a)\textbf{(a)} We know that the formula for mean xˉ\bar{x} with score value xix_i and total items nn is,

xˉ=xin\bar{x}=\dfrac{\sum x_i}{n}

This implies that the mean value for the given weight data would be,

xˉ=0.8+1.1+1.2+0.9+1.2+1.2+0.9+0.7+1.0+1.110xˉ=10.110xˉ=1.01\begin{align*} \bar{x} &= \dfrac{0.8+1.1+1.2+0.9+1.2+1.2+0.9+0.7+1.0+1.1}{10}\\\\ \bar{x} &= \dfrac{10.1}{10}\\\\ \bar{x} &= 1.01\\ \end{align*}

Next, there are two ways to calculate the standard deviation.

Case 1: If we consider the given sample as whole population then the formula for the standard deviation ss with score value xix_i and mean xˉ\bar{x} is,

s=(xixˉ)2ns=\sqrt{\dfrac{\sum (x_i - \bar{x})^2}{n}}

Case 2: When we use the sample as an estimate of the whole population then the formula for the standard deviation ss with score value xix_i and mean xˉ\bar{x} is,

s=(xixˉ)2n1s=\sqrt{\dfrac{\sum (x_i - \bar{x})^2}{n - 1}}

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