Q R1.6. - The Practice of Statistics for AP

Question

The 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:

(a) Present these measurements graphically in a stem plot.

(b) Discuss the shape, center, and spread of the distribution. Are there any outliers?

(c) What is your estimate of the density of the earth based on these measurements? Explain.

Step-by-Step Solution

Answer

Part (b) No. 

Part (c) The estimate is $5.4479$

Part (a) The data set is 

48	8
49
50	7
51	0
52	6	7	9	9
53	0	4	4	6	9
54	2	4	6	7
55	0	3	5	7	8
56	1	2	3	5	8
57	5	9
58	5

Part (a) Step 1: Given information

The following are the $29$ measurements of the earth's density as a multiple of the density of water made by Cavendish: 

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

Part (a) Step 2: Concept

The formula to compute the mean is: $Mean = \frac{Sum of observations}{Number of observations}$

Part (a) Step 3: Explanation

The stem and leaf plot is constructed as: 

48	8
49
50	7
51	0
52	6	7	9	9
53	0	4	4	6	9
54	2	4	6	7
55	0	3	5	7	8
56	1	2	3	5	8
57	5	9
58	5

Part (b) Step 1: Explanation

Because most of the data is in the center of the plot, the above-constructed plot reveals that the distribution is approximately symmetric. 

Center: At the highest bar, the center has a value of $5.5$

Data has taken values ranging from $4.88$ to $5.85$

There are no values that are far apart from each other. As a result, this distribution does not contain any outliers. 

Part (c) Step 1: Calculation

The density of the earth can be estimated using the computed sample mean as follows: $$\begin{gathered} \overline{x}=\frac{\Sigma x}{n} \\ = \frac{5.50+5.61+........+5.85}{29} \\ = \frac{157.99}{29} \\ = 5.4479 \end{gathered}$$

Thus, the required value is $5.4479$