Density of the earth In 1798, the English scientist Henry Cavendish measured the density of the earth several times by careful work with a torsion balance. The variable recorded was the density of the earth as a multiple of the density of water. Here are Cavendish’s 29 measurements:

Part (a). Make a stemplot of the data.

Part (b). Describe the distribution of density measurements.

Part (c). The currently accepted value for the density of earth is 5.51 times the density of water. How does this value compare to the mean of the distribution of density measurements?

5.50, 5.61, 4.88, 5.07, 5.26, 5.55, 5.36, 5.29, 5.58, 5.65, 5.57, 5.53, 5.62, 5.29, 5.44, 5.34, 5.79, 5.10, 5.27, 5.39, 5.42, 5.47, 5.63, 5.34, 5.46, 5.30, 5.75, 5.68, 5.85

Sort the following data values from smallest to largest:

4.88, 5.07, 5.1, 5.26, 5.27, 5.29, 5.29, 5.3, 5.34, 5.34, 5.36, 5.39, 5.42, 5.44, 5.46, 5.47, 5.5, 5.53, 5.55, 5.57, 5.58, 5.61, 5.62, 5.63, 5.65, 5.68, 5.75, 5.79, 5.85

Stemplot

Place the tenths to the left of the vertical line and the hundredths to the right of the vertical line for each data value.

Key: $\overline{)50}7$mean 5.07

Shape: We can see that the distribution is roughly symmetric because the majority of the values are in the middle (when ignoring the outlier 4.88).

The median is the value in the middle of the sorted data set. Because there are 29 outliers in the data set, the median is the 15th data value in the sorted data set, which is 5.46.

The data values range from 4.88 to 5.85, as we can see.

Unusual features: 4.88 is a possible outlier because it is separated from the other data values in the stemplot by a gap.

As a result:

The centre of the roughly symmetric distribution is at 5.46 (median) and the data values range from 4.88 to 5.85.

4.88 could be an outlier.

We can see that the given data set has 29 different data values.

n = 29

The sum of all values divided by the number of values yields the mean:

$\overline{x}=\frac{\sum _{i=1}^n{x}_i}{n}$

$$\begin{gathered} \overline{x}=\frac{5.50+5.61+4.88+5.07+5.26+5.55+5.36+5.29+5.58+5.65+5.57+5.53+5.62+5.29+5.44+5.34+5.79+5.10+5.27+5.39+5.42+5.47+5.63+5.34+5.46+5.30+5.75+5.68+5.85}{29} \\ \overline{x}=\frac{157.99}{29} \\ \overline{x}\approx 5.4479 \end{gathered}$$

The mean of the distribution (5.4479) is then slightly lower than the value 5.51.