By:

Eric N. Madan

Arqam Iqbal Pasha

CCT 226: PRESENTATION

Joke!

A statistician can have his head in an oven and his feet in ice, and he will say that on the average he feels fine.

QUESTION 108(a)

Refer to the real estate data from the Central Area of Edmonton, which reports information on listed homes.

a. Select the variable size (Sq. Ft.) 1. Find the mean, median and the standard

deviation.2. Determine the coefficient of skewness, is the

distribution positively or negatively skewed?3. Develop a box plot. Are there any outliers?

Estimate the first and third quartiles.4. Summarize the results.

*unless you are sure of what you are doing!

Data Set

Class Distribution Mid-Points (X) Frequency (f) f * X X^2 f * (X^2)

672 - 997 834.50 19 15,855.50 696,390.25 581,137,663.63

998 - 1323 1,060.50 34 36,057.00 1,124,660.25 1,192,702,195.13

1324 - 1649 1,486.50 15 22,297.50 2,209,682.25 3,284,692,664.63

1650 - 1975 1,812.50 10 18,125.00 3,285,156.25 5,954,345,703.13

1976 - 2301 2,138.50 12 25,662.00 4,573,182.25 9,779,750,241.63

2302 - 2627 2,464.50 4 9,858.00 6,073,760.25 14,968,782,136.13

2628 - 2953 2,790.50 2 5,581.00 7,786,890.25 21,729,317,242.63

    96 133,436.00   57,490,727,846.88

Result

Mean = 1,389.96 Median = 1239.5Mode = 1119

Standard Deviation = 24560.39974Coefficient of Skewness = 0.018378162

Let’s Get Serious

Arithmetic Mean

n

iiX

NMean

1

1

Arithmetic Mean (cont.)

1.414,196

754,135Mean

Median

thMedian

thN

Median

2

196

2

1

Median = 48.5th term = 1239.5

Mode

1119Mode

Standard Deviation

1

2

2

nn

XX

s

568.497s

Coefficient of Skewness

s

MedianMeansk

3

568.497

5.239,11.414,13 sk

Coefficient of Skewness (cont.)

93816.0skewness

First Quartile

L25 = 24.25th term

P25 = 1050.25

4

1)196(

4

11

25

25

L

nL

Third Quartile

L75 = 72.75th term

P75 = 1793

4

3)196(

4

31

75

75

L

nL

Box Plot

There is one outlier = 2947

Summary

Mean = 1,414.10 Median = 1239.5Mode = 1119

Standard Deviation = 497.568Coefficient of Skewness = 0.93816

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