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## Southern New Hampshire University Weather Application of Probability Discussion & Worksheet

### Question Description

Using Option 1 discussion, respond to Student 2 (listed below) Use the discussion topic (Option 1 below) as a place to ask questions, speculate about answers, and share insights. Be sure to embed and cite your references for any supporting images.

Option 1:

Given this data set – from the NOAA for Manchester, NH, select a random month between January 1930 and December 1957. Begin with this month and analyze the next 25 data values (i.e. 2 years and 1 month) for the variable “TPCP” (See Second Tab in Data Set for variable descriptions). For example, if May 1955 is chosen as the starting month, then the “TPCP” data would be from May 1955 through May 1957. Using Excel, StatCrunch, etc., construct a histogram to represent your sample. Report the sample mean, median, and standard deviation as a part of your discussion of skewness.

Determine the interval for the middle 68% of your sample data and relate this to the sample standard deviation.

Comment on the similarities and differences between your sample data and that of your classmates. Why are there differences if the samples are drawn from the same population?

STUDENT 2: Respond to this students post with no plagiarism (all work is ran through a plagiarism site). Has to be substantial response. No word limit.

I chose option 1 for this week’s discussion. I must say it was a bit challenging and I hope I was able to input the numbers correctly using Statcrunch. The random month I chose was January 1951 through January 1953. Below you will find the sample mean, median, standard deviation and histogram chart.

Sample Mean: 1055.8

Median: 1000

Standard Deviation: 406.68

Histogram:

The skewness of the graph I created resembles that of Bell-Shaped. The interval for 68% begins with the bar that reaches a frequency of 7 followed by the bar next to it that reaches a frequency of 5 and lastly the bar that reaches a frequency of 6, which ranges from 750 to 1500.

Using the Empirical Rule 750-1055.8(mean) divided by 406.68(standard deviation) gave me -0.75. So 750 is -0.75 standard deviation below the mean.

I then did the same for 1500. 1500- 1055.8(mean) divided by 406.68(standard deviation) gave me 1.09. So 1500 is 1.09 away from the mean.

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