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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鈥檚 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.

Short Answer

Expert verified

Part (b) No.

Part (c) The estimate is 5.4479

Part (a) The data set is

488



49




507



510



526799
5304469
542467
5503578
5612358
5759


585



Step by step solution

01

Part (a) Step 1: Given information

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

5.505.614.885.075.265.555.365.295.585.65
5.575.535.625.295.445.345.795.105.275.39
5.425.475.635.345.465.305.755.685.85
02

Part (a) Step 2: Concept

The formula to compute the mean is: Mean=SumofobservationsNumberofobservations

03

Part (a) Step 3: Explanation

The stem and leaf plot is constructed as:

488



49




507



510



526799
5304469
542467
5503578
5612358
5759


585



04

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.

05

Part (c) Step 1: Calculation

The density of the earth can be estimated using the computed sample mean as follows: x=xn=5.50+5.61+........+5.8529=157.9929=5.4479

Thus, the required value is 5.4479

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