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Displaying Data in a Histogram

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I need help in how to display the following data in a histogram.

Can you provide step by step advice on how I can go about solving this problem?

1. Display the frequency distribution of the salaries graphically in a histogram (choose intervals to use along the horizontal axis) and tell what is the mean, median and skew of the distribution.

The High school Basket ball team salaries for 2005 are listed below:

7.88M 7.70M 4.20M 6.60M 6.19M 920K

5.37M 1.35M 870K 760K 950K 4.16M

2.65M 1.97M 5.22M 720K 1.61M 720K

3.55M 1.65M 1.58M 810K 850K 470K

600K 590K 340K 510K 470K 850K

400K 600K 390K 450K 370K 1.66M

280K 950K 990K 720K 1.30M 410K

580K 300K 300K 440K 270K 540K

960K 460K 290K

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Solution Preview

I simply entered the data into excel then wrote everything in thousands so 500k = 500 and 5M = 5000. I then used Excel's histogram tool. I ...

Solution Summary

Using the power of Excel, this solution shows how to display a frequency distribution histogram for basket ball team salary data.

See Also This Related BrainMass Solution

Use histograms to display the distribution - Comment on the symmetry; is it skewed?

Compare J&J and Merck stock to determine which stock will provide the investor with the best return for their money. Stock data was collected for 30 days between November 30th, 2006 and January 10th, 2007. The measures of central tendency and dispersion of the collected data will be determined and examined to complete the analysis. First, the measures of central tendency and dispersion will be calculated. The descriptive statistical data will then be displayed using graphical and tabular techniques. The distribution will be displayed using histograms. The symmetry will be examined to determine if it is skewed and the data will be explained referencing the display. The measures that are most appropriate to draw conclusions from based on the parameters of the stock numbers will be determined. By completing the statistical equations and displaying the data, the team will be able to determine whether J&J or Merck is the better investment based on the 30 days of collected data. The chart below displays the 30 data points for each organization.
11/30/2006 65.91 44.51
12/1/2006 65.97 45.06
12/4/2006 66.28 44.7
12/5/2006 66.17 45.39
12/6/2006 66.03 44.67
12/7/2006 66.06 43.95
12/8/2006 65.95 43.93
12/11/2006 65.7 44.01
12/12/2006 65.58 43.62
12/13/2006 65.47 43.34
12/14/2006 66.25 43.6
12/15/2006 66.29 44
12/18/2006 66.6 43.83
12/19/2006 66.88 43.51
12/20/2006 66.43 43.3
12/21/2006 66 43.27
12/22/2006 65.65 42.83
12/25/2006 65.65 42.83
12/26/2006 65.64 42.79
12/27/2006 66.01 43.37
12/28/2006 66.42 43.55
12/29/2006 66.02 43.6
1/1/2007 66.02 43.6
1/2/2007 66.02 43.6
1/3/2007 66.4 44.02
1/4/2007 67.23 45.11
1/5/2007 66.62 44.3
1/8/2007 66.51 44.29
1/9/2007 66.26 43.88
1/10/2007 66.15 44.12

Measures of Central Tendency and Dispersion
The measure of central tendency includes the mean, median, and mode of a set of variables. Thirty days of data points were collected for J&J stock and the same 30 days were collected for Merck stock. The mean or average stock value for J&J during the 30 day period was 66.139, the Merck stock was 43.886. The median or mid range of the J&J stock was 66.045, the Merck stock was 43.855. The mode or the most frequent variable in the J&J data was 66.02, and Merck was 43.6 as these values appeared the most in each set of data.
Ways to measure the dispersion include the variance, standard deviation, range, and interquartile range. The population standard deviation is used and not the sample because all of the data points are known. The variance for J&J stock is .150556 and Merck is .420698. The standard deviation is the square root of the variance and for the J&J 30 points of stock is .3990150341, Merck was .6486118511. The range is 1.76 as was determined by subtracting the lowest variable, 65.47 from the highest variable of 67.23 for J&J and Merck was 2.6 by subtracting 42.79 from 45.39. The first quartile range for J&J is 65.95 and is the 8th variable in the dataset. The median is the middle of the data set and since 30 is an even number; it falls between the middle two numbers of the data set and is 66.045. The 3rd quartile is 66.4 and is the 23rd number or the 8th from the right of the dataset. The interquartile range is .45 and is determined by subtracting the Q1 65.95 from the Q3, 66.4. The first quartile for Merck is 43.51, the third quartile is 44.29 and the interquartile range is .78. The chart below displays the central tendency and dispersion for both J&J and Merck.
J&J & Merck Central Tendency & Dispersion

J&J Merck

Data Points 30 30
Mean 66.139 43.886
Median 66.045 43.855
Mode 66.02 43.6
Variance 0.150556 0.420698
Standard deviation (Population 0.388015 0.648612
Min 65.47 42.79
Max 67.23 45.39
Range 1.76 2.6
1st Quartile 65.95 43.51
3rd Quartile 66.4 44.29
Interquartile Range 0.45 0.78

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