Question

below are the lengths of 10 randomly selected newborns and 10 randomly selected six - month - old babies. newborn lengths (inches): 18, 19, 19, 19.5, 20, 20, 20, 20.5, 20.5, 21. study the data given. was there more variation in the lengths of newborns or in the lengths of six - month - olds? six - month - olds. newborns. both showed the same amount of variation in length

Explanation:

Step1: Calculate variance of newborn lengths

First, find the mean of newborn lengths: $\bar{x}_{newborn}=\frac{18 + 19+19+19.5+20+20+20+20.5+20.5+21}{10}=\frac{197.5}{10} = 19.75$. Then calculate the variance $s_{newborn}^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$.
$(18 - 19.75)^2+(19 - 19.75)^2+(19 - 19.75)^2+(19.5 - 19.75)^2+(20 - 19.75)^2+(20 - 19.75)^2+(20 - 19.75)^2+(20.5 - 19.75)^2+(20.5 - 19.75)^2+(21 - 19.75)^2$
$=(-1.75)^2+(-0.75)^2+(-0.75)^2+(-0.25)^2+(0.25)^2+(0.25)^2+(0.25)^2+(0.75)^2+(0.75)^2+(1.25)^2$
$=3.0625 + 0.5625+0.5625+0.0625+0.0625+0.0625+0.0625+0.5625+0.5625+1.5625 = 7.25$.
$s_{newborn}^2=\frac{7.25}{9}\approx0.806$.

Step2: Assume we have data for six - month - olds

Since the data for six - month - olds is not given in the question, we cannot calculate its variance. But if we assume we had the data, we would follow the same process as in Step 1: find the mean $\bar{x}_{six - month}$, then calculate $(x_i-\bar{x}_{six - month})^2$ for each data point, sum them up and divide by $n-1$ (where $n$ is the number of six - month - old data points).
However, since we are missing the six - month - old data, we cannot make a valid comparison. But if we assume we had the data and calculated the variance of six - month - olds $s_{six - month}^2$, we would compare $s_{newborn}^2$ and $s_{six - month}^2$. If $s_{newborn}^2>s_{six - month}^2$, there is more variation in newborns, if $s_{newborn}^2

Since the data for six - month - olds is missing, we cannot answer this question.

Answer:

Insufficient data to answer.