Answer:
The standard deviation for week two was about 3 ounces more than the standard deviation for week one
Step-by-step explanation:
Given
[tex]Week\ 1: 128, 105, 80, 82, 96, 98, 87, 100, 112, 126[/tex]
[tex]Week\ 2: 75, 85, 90, 97, 89, 105, 110, 127, 129, 130[/tex]
See attachment for options
Required
The true statement
Checking the standard deviation
For week 1
Calculate the mean:
[tex]\bar x = \frac{\sum x}{n}[/tex]
[tex]\bar x = \frac{128+ 105+ 80+ 82+ 96+ 98+ 87+ 100+ 112+ 126}{10}[/tex]
[tex]\bar x = \frac{1014}{10}[/tex]
[tex]\bar x_1 = 101.4[/tex]
Then standard deviation
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{(128 - 101.4)^2 +............+ (126- 101.4)^2}{10}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{2522.4}{10}}[/tex]
[tex]\sigma_1 = \sqrt{252.24[/tex]
[tex]\sigma_1 = 15.88[/tex]
For week 2, we have:
[tex]\bar x = \frac{75+ 85+ 90+ 97+ 89+ 105+ 110+ 127+ 129+ 130}{10}[/tex]
[tex]\bar x = \frac{1037}{10}[/tex]
[tex]\bar x_2 = 103.7[/tex]
Then standard deviation
[tex]\sigma_2 = \sqrt{\frac{(75 - 103.7)^2 +................+ (130- 103.7)^2}{10}}[/tex]
[tex]\sigma_2 = \sqrt{\frac{3538.1}{10}}[/tex]
[tex]\sigma_2 = \sqrt{353.81[/tex]
[tex]\sigma_2 = 18.81[/tex]
Compare the standard deviations
[tex]\sigma_1 = 15.88[/tex]
[tex]\sigma_2 = 18.81[/tex]
Calculate the difference:
[tex]d = \sigma_2 - \sigma_1[/tex]
[tex]d = 18.81 - 15.88[/tex]
[tex]d = 2.93[/tex]
[tex]d \approx 3[/tex]
This implies that option (b) is true
