Question

you work for a large farm with many fields of corn. you are investigating the mass in grams of a sample of ears of corn. you gather the following data: mass(g) 455.1 585.6 601.8 545.5 519.8 756.5 494.7 677.8 464.6 471.8 634 470 507.6 some of the masses in the sample seem much larger than the rest. you decide to make several calculations to help in the search for outliers. report your answer with all the decimal places. find the following: a. the 26th percentile = b. iqr = c. apply the 1.5 iqr rule to search for outliers. report the lower and upper cutoffs limits (fences).

Explanation:

Step1: Sort the data

First, sort the data in ascending - order: 455.1, 464.6, 470, 471.8, 494.7, 507.6, 519.8, 545.5, 585.6, 601.8, 634, 677.8, 756.5. There are \(n = 13\) data points.

Step2: Calculate the 26th percentile

The index \(i\) for the \(p\)th percentile is given by \(i=\frac{p}{100}\times n\). For \(p = 26\) and \(n = 13\), \(i=\frac{26}{100}\times13=3.38\). Since \(i\) is not an integer, we round up to the next - integer. So the 26th percentile is the 4th value in the sorted data set, which is 471.8.

Step3: Calculate the first and third quartiles

The first quartile \(Q_1\): The index for \(Q_1\) (25th percentile) is \(i=\frac{25}{100}\times13 = 3.25\). Rounding up, \(Q_1\) is the 4th value in the sorted data set, \(Q_1 = 471.8\).
The third quartile \(Q_3\): The index for \(Q_3\) (75th percentile) is \(i=\frac{75}{100}\times13=9.75\). Rounding up, \(Q_3\) is the 10th value in the sorted data set, \(Q_3 = 601.8\).
The inter - quartile range \(IQR=Q_3 - Q_1\), so \(IQR=601.8−471.8 = 130\).

Step4: Calculate the lower and upper fences

The lower fence \(=Q_1-1.5\times IQR\). Substitute \(Q_1 = 471.8\) and \(IQR = 130\), we get \(471.8-1.5\times130=471.8 - 195=276.8\).
The upper fence \(=Q_3 + 1.5\times IQR\). Substitute \(Q_3 = 601.8\) and \(IQR = 130\), we get \(601.8+1.5\times130=601.8 + 195=796.8\).

Answer:

a. 471.8
b. 130
c. Lower cutoff limit: 276.8, Upper cutoff limit: 796.8