Question

A hospital recorded the weights, in ounces, of newborn babies for two weeks. The

Answer 1

Answer:

The standard deviation for week two was about 3 ounces more than the standard deviation for week one

Step-by-step explanation:

Given

$Week\ 1: 128, 105, 80, 82, 96, 98, 87, 100, 112, 126$

$Week\ 2: 75, 85, 90, 97, 89, 105, 110, 127, 129, 130$

See attachment for options

Required

The true statement

Checking the standard deviation

For week 1

Calculate the mean:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{128+ 105+ 80+ 82+ 96+ 98+ 87+ 100+ 112+ 126}{10}$

$\bar x = \frac{1014}{10}$

$\bar x_1 = 101.4$

Then standard deviation

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

$\sigma_1 = \sqrt{\frac{(128 - 101.4)^2 +............+ (126- 101.4)^2}{10}}$

$\sigma_1 = \sqrt{\frac{2522.4}{10}}$

$\sigma_1 = \sqrt{252.24$

$\sigma_1 = 15.88$

For week 2, we have:

$\bar x = \frac{75+ 85+ 90+ 97+ 89+ 105+ 110+ 127+ 129+ 130}{10}$

$\bar x = \frac{1037}{10}$

$\bar x_2 = 103.7$

Then standard deviation

$\sigma_2 = \sqrt{\frac{(75 - 103.7)^2 +................+ (130- 103.7)^2}{10}}$

$\sigma_2 = \sqrt{\frac{3538.1}{10}}$

$\sigma_2 = \sqrt{353.81$

$\sigma_2 = 18.81$

Compare the standard deviations

$\sigma_1 = 15.88$

$\sigma_2 = 18.81$

Calculate the difference:

$d = \sigma_2 - \sigma_1$

$d = 18.81 - 15.88$

$d = 2.93$

$d \approx 3$

This implies that option (b) is true