If finding the standard deviation is defined as the square root of the variance, adding and subtracting numbers in the data will not affect the variance, because the distance between values does not change.
- a) To find the mean and the standard deviation if I multiplied each height by 5.
I would first have to multiply 5 to each height which is shown in the table below:
Then repeating steps I used on the first question using my TI-84 (STAT: CALC: 1-Var Stats: ENTER: List: 2ND: 2: ENTER: Calculate: ENTER). The following data achieved is shown below:
The mean and standard deviation of multiplying each height by 5 would be 764.5833333 as the mean, and 85.43658304 as the standard deviation.
b) To find the mean and the standard deviation if I multiplied each height by 0.2.
I would first have to multiply each height by 0.2 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). The results shown below:
The mean and standard deviation of multiplying each height by 0.2 would be 30.58333333 as the mean, and 3.417463322 as the standard deviation.
If you were to multiply every value by a constant, then the mean will also be multiplied by that same constant. For example, the results achieved from question one had a mean of 152.9166667. If we multiply each of those scores by 5, the new mean will be 152.9166667 × 5= 764.5833335 as a result of question 2. If you were to multiply every score by that same constant, then the standard deviation will be affected because it has a greater effect on the variance. If the constant is smaller than 0, the mean and standard deviation would be a negative number.
- The table below is a transformation of the values in the information given into a cumulative frequency table.
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.
- 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) To find the median and the interquartile range if I subtracted 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 (STAT: CALC: 1-Var Stats: ENTER: List: 2ND: 3: 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 subtracting 12 cm to each height gives you 136.5 as the median, and 33 as the interquartile range.
If finding the median is to find the number in the middle of values, it would affect it because you are adding and subtracting numbers from every value, therefore the mean increased and decreased by the same constant just like the mean. If finding the interquartile range is defined as The distance between the top of the lower quartile and the bottom of the upper quartile of a given value adding and subtracting numbers in the data will not affect the variance, because the distance between values does not change.
- a) To find the median if I multiplied each height by 5, I would first have to multiply each height by 5, which is shown in the table below:
Then I would repeat 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
To find the interquartile range, I would have to use Q3-Q1 = 850-685, it should give me 165. The median and interquartile range of multiplying each height by 5 gives you 742.5 as the median, and 165 as the interquartile range.
b) To find the mean and the standard deviation if I multiplied each height by 0.2. I would first have to multiply each height by 0.2 which is shown in the table below:
Then, I would repeat 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:
To find the interquartile range, I would have to use Q3-Q1= 175-142, which should give me 6.6. The median and interquartile range of multiplying each height by 0.2, gives you 29.7 as the median, and 6.6 as the interquartile range.
If you were to multiply every value by a constant, then the median will also be multiplied by that same constant. For example, the results we had from question one had a median of 148.5. If we multiply each of those scores by 5, the new mean will be 148.5 × 5= 742.5 as a result of question 6. This is just like the mean. If you were to multiply every score by that same constant, then the interquartile range will be affected because it has a greater effect on the variance. This is just like standard deviation. If you’re constant is smaller than 0 your number would become a negative median and interquartile range.
- 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 the interquartile range and standard deviation this will cause a greater effect on the variance.
- 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)