Question

Randomly select 10 values from the number of suspensions in the local school districts in southwestern Pennsylvania in Data Set V in Appendix B. Find the mean, median, mode, range, variance, and standard deviation of the number of suspensions by using the Pearson coefficient of skewness.

 

Data from Set V Appendix B

Answer 1

The number of suspensions of the first 10 people in data bank is as follows:

37, 29, 106, 47, 51, 46, 65, 223, 10, 60

 

The data set is increasing order is as follows:

10, 29, 37, 46, 47, 51, 60, 65, 106, 223

 

Mean,

X̅ = ΣX/n

    = (10 + 29 + 37 + 46 + 47 + 51 + 60 + 65 + 106 + 223)/10

   = 674/10

   = 67.4

 

Since the middle point falls halfway between 5th and 6th, find the median MD by adding the two values and dividing by 2.

 

Median = (5th data value + 6th data value)/2

               = (47 + 51)/2

               = 98/2

               = 49

 

There is no mode as each data value appears exactly once.

Range = Highest value – Lowest value

            = 223 – 10

           = 213

 

Sum of the squares of the given data set is,

ΣX2 = 102 + 292 + 372 + 462 + 472 + 512 +602 + 652 + 1062 + 2232

        = 100 + 841 + 1369 + 2116 + 2209 + 2601 + 3600 + 4225 + 11236 + 49729

        = 78026

 

So, the variance for the given data set is,

s2 = [ΣX2 – {(ΣX)2/n}]/(n – 1)

     = [78026 – (67.4)2/10]/9

     = (78026 – 45427.6)/9

     = 32598.4/9

     = 3622

 

The standard deviation for the given data set is

Standard deviation,

s = √variance

   = √3622

   = 60.2

Pearson coefficient of skew ness is as follows:

Skewness = 3(X̅ - MD)/s

                   = 3(67.4 – 49)/60.2

                   = 0.92

 

Since the Pearson coefficient of skew ness is positive, the distribution is positively skewed.

Since the Pearson coefficient of skew ness is positive, the distribution is positively skewed.