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# This assignments purpose is to investigate how translation and enlargement of data affects statistical parametersI

Extracts from this document...

Introduction

3 Transforming Data Type I

This assignment’s purpose is to investigate how translation and enlargement of data affects statistical parameters.

The following data I entered in a TI-84 using STAT: EDIT: ENTER

This table shows the height in centimeters of 60 students.

1. To find the mean (The sum of all observed values divided by the total number of values.) and the Standard deviation (The most used of all measures of dispersion. Simply defined as the square root of the variance.), I used my TI-84 (STAT: CALC: 1-Var Stats: ENTER: List: 2ND: 2: ENTER: Calculate: ENTER), which is shown the picture below

The mean of the student’s hieght is 152.9166667 and the standard deviation is 17.08731661.

1.  a) To find the mean and the standard deviation, if I added 5 cm to each height I would first have to add 5 cm to each height which is shown in the table below:

Then, repeating the steps I used on the first question using my TI-84 (STAT: CALC: 1-Var Stats: ENTER: List: 2ND: 2: ENTER: Calculate: ENTER) which gives me the following data:

The mean and standard deviation of adding 5 cm to each height gives you 157.9166667 as the mean, and 17.08731661 as the standard deviation.

b) To find the mean and the standard deviation if subtracting 12 cm from each height, I would first have to subtract 12 cm from each height which is shown in the table below:

Then I would repeat the steps I used on the first question using my TI-84

Middle

1. The table below is a transformation of the values in the information given into a cumulative frequency table.
 Height (cm) Frequency CumulativeFrequency
 130-139 17 17
 140-149 14 31
 150-159 5 36
 160-169 9 45
 170-179 15 60

The graph below would represents this data in a cumulative frequency graph

We first would need to find the median; the median is the middle number of the values. If we find the 50th percentile, it should give us 148.5. Then to find the interquartile range we would have to find the lower quartile and the higher quartile which is Q1=137, Q3=170, after we would use the formula to find the interquartile range (Q3-Q1) which should give us 33 as the interquartile range.

1. a) To find the median and the interquartile range if 5 cm was added to each height, I would first have to add 5 cm to each height which is shown in the table below:

Then repeating steps used in the first question, using my TI-84 (STAT: CALC: 1-Var Stats: ENTER: List: 2ND: 2: ENTER: Calculate: ENTER) which gives me the following data:

To find the interquartile range I would have to use Q3-Q1= 175-142, which should give me 33. The median and interquartile range of adding 5 cm to each height gives you 153.5 as the median, and 33 as the interquartile range.

b)

Conclusion

1. My results have proven that the mean and median are quite similar; it also applies to the interquartile range and standard deviation. If you add or subtract a constant to every value, the mean and median will increase or decrease according to the constant. As for the interquartile range and standard deviation, the distance value has not been changed, and all the variability’s remain the same.

If we were to multiply every value by that same constant, the mean and median will also be multiplied by that constant. As for theinterquartile range and standard deviation this will cause a greater effect on the variance.

1. a) If I were to transform my given data so that the mean would be 0, I would have to multiply every value by a constant of 0.

b) If I were to transform my given data so that the standard deviation would be 1, I would have to use algebra to find what I would do to multiply my standard deviation so it would be 1:

17.08731661 × n = 1

17.08731661 × n  = _      1           _

17.08731661            17.08731661

n = _           1        _

17.08731661

n = 0.0585229397

c)

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