Question

a) compute the mean and standard deviation of the distribution.
the mean is 8.
the standard deviation is 4.85.
(round to the nearest hundredth as needed.)
b) multiply each number in the distribution by 2 and compute the mean and the standard deviation of this new distribution.
the mean is 16.
the standard deviation is 9.7.
(round to the nearest hundredth as needed.)
c) multiply each number in the original distribution by 4 and compute the mean and the standard deviation of this new distribution.
the mean is 32.
the standard deviation is 19.4.
(round to the nearest hundredth as needed.)
d) the mean and standard deviation of the distribution 1, 3, 4, 4, 5, 7 are 4 and 2, respectively. draw a conclusion about how the mean and standard deviation change when each value in a distribution is multiplied by the same number, and use this conclusion to determine the mean and standard deviation of the distribution 2, 6, 8, 8, 10, 14.
the mean is

Explanation:

Step1: Recall the property of mean and standard deviation under scalar multiplication

If \(x_1,x_2,\cdots,x_n\) is a data - set with mean \(\bar{x}\) and standard deviation \(s\), and we create a new data - set \(y_i = kx_i\) for \(i = 1,2,\cdots,n\) (where \(k\) is a scalar), then the mean of the new data - set \(\bar{y}=k\bar{x}\) and the standard deviation of the new data - set \(s_y = k s_x\).

Step2: Analyze part (d)

In part (d), the original data - set (let's assume it's data - set 1: \(1,3,4,4,5,7\)) has mean \(\bar{x}_1\) and standard deviation \(s_1\). The new data - set (data - set 2: \(2,6,8,8,10,14\)) is obtained by multiplying each element of data - set 1 by \(k = 2\).
We know from part (a) (mean of data - set 1: \(\bar{x}_1=4\), standard deviation of data - set 1: \(s_1 = 2\)) and part (b) (when \(k = 2\), \(\bar{y}=k\bar{x}=2\times4 = 8\), \(s_y=k s_x=2\times2=4\)), part (c) (when \(k = 4\), if the original mean is \(\bar{x}\) and standard deviation is \(s\), \(\bar{y}=k\bar{x}=4\times4 = 16\), \(s_y=k s_x=4\times2 = 8\)).
For data - set 2 (which is \(k = 2\) times data - set 1), using the formula \(\bar{y}=k\bar{x}\) and \(s_y=k s_x\). If we consider the general formula, when we multiply each data point in a data - set by a constant \(k\), the mean and the standard deviation are also multiplied by \(k\).

Let the original data - set be \(x_1,x_2,\cdots,x_n\). The mean \(\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i\). For the new data - set \(y_i=kx_i\), the mean \(\bar{y}=\frac{1}{n}\sum_{i = 1}^{n}y_i=\frac{1}{n}\sum_{i = 1}^{n}(kx_i)=k\frac{1}{n}\sum_{i = 1}^{n}x_i=k\bar{x}\).

The formula for the standard deviation \(s=\sqrt{\frac{1}{n}\sum_{i = 1}^{n}(x_i-\bar{x})^2}\). For the new data - set \(y_i = kx_i\) and \(\bar{y}=k\bar{x}\), \(s_y=\sqrt{\frac{1}{n}\sum_{i = 1}^{n}(y_i - \bar{y})^2}=\sqrt{\frac{1}{n}\sum_{i = 1}^{n}(kx_i - k\bar{x})^2}=\sqrt{\frac{1}{n}\sum_{i = 1}^{n}k^2(x_i - \bar{x})^2}=k\sqrt{\frac{1}{n}\sum_{i = 1}^{n}(x_i - \bar{x})^2}=k s\).

Since data - set 2 is obtained by multiplying data - set 1 by \(k = 2\), if the mean of data - set 1 is \(4\) and standard deviation is \(2\), then for data - set 2, the mean is \(2\times4=8\) and the standard deviation is \(2\times2 = 4\).

Answer:

The mean is \(8\) and the standard deviation is \(4\).