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

Explanation:

Step1: Recall the formula for mean and standard deviation when data is scaled

If \(x_1,x_2,\cdots,x_n\) is a data - set with mean \(\bar{x}\) and standard deviation \(s\), and \(y_i = kx_i\) (where \(k\) is a constant), then the mean of the new data - set \(y_1,y_2,\cdots,y_n\) is \(\bar{y}=k\bar{x}\) and the standard deviation is \(s_y = k s_x\).

Step2: Solve for part (b)

Given \(k = 2\), \(\bar{x}=8\), \(s_x = 4.85\)

• For the mean: \(\bar{y}=k\bar{x}\)

Substitute \(k = 2\) and \(\bar{x}=8\) into the formula, we get \(\bar{y}=2\times8=16\)

• For the standard deviation: \(s_y=k s_x\)

Substitute \(k = 2\) and \(s_x = 4.85\) into the formula, \(s_y=2\times4.85 = 9.7\)

Step3: Solve for part (c)

Given \(k = 4\), \(\bar{x}=8\), \(s_x = 4.85\)

• For the mean: \(\bar{y}=k\bar{x}\)

Substitute \(k = 4\) and \(\bar{x}=8\) into the formula, \(\bar{y}=4\times8 = 32\)

• For the standard deviation: \(s_y=k s_x\)

Substitute \(k = 4\) and \(s_x = 4.85\) into the formula, \(s_y=4\times4.85=19.4\)

Answer:

b) The mean is \(16\) and the standard deviation is \(9.7\).
c) The mean is \(32\) and the standard deviation is \(19.4\).