Solved Examples: Measures of Position

Student $E$ made the highest test score because the score is $2.00$ standard deviations above the mean.

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \underline{Agnes'\:\:Chemistry\:\:Test} \\[3ex] x = 73 \\[3ex] \mu = 95 \\[3ex] \sigma = 11 \\[3ex] z = \dfrac{73 - 95}{11} \\[5ex] z = -\dfrac{22}{11} \\[5ex] z = -2.00 \\[3ex] \underline{Agatha's\:\:Biology\:\:Test} \\[3ex] x = 46 \\[3ex] \mu = 65 \\[3ex] \sigma = 10 \\[3ex] z = \dfrac{46 - 65}{10} \\[5ex] z = -\dfrac{19}{10} \\[5ex] z = -1.90 \\[3ex] -1.90 \gt -2.00 \\[3ex] $ Therefore, Agatha's Biology test score is better because it has a higher $z-score$ than Agnes' Chemistry test score.

Note the responses of your students.
If some of them said that Agnes' test score was better, re-teach the concept of $z-scores$

(a.) A data score is usual if $-2.00 \le z \le 2.00$
A data value is unusual if the $z-score \lt -2.00$ OR the $z-score \gt 2.00$
The lower $z-score$ boundary is $-2.00$
The upper $z-score$ boundary is $2.00$
This implies that $-2.00$ and $2.00$ are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores

(b.) To find the $IQ$ scores (data values), we need to express them in terms of the $z-scores$.

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \rightarrow x - \mu = z\sigma \\[3ex] x = z\sigma + \mu \\[3ex] Lower\:\:IQ\:\:score\:\:boundary = -2(19) + 107 = -38 + 107 = 69 \\[3ex] Upper\:\:IQ\:\:score\:\:boundary = 2(19) + 107 = 38 + 107 = 145 \\[3ex] Usual\:\:IQ\:\:scores:\:\: 69 \le IQ \le 145 \\[3ex] Unusual\:\:IQ\:\:scores:\:\: IQ \lt 69 \:\:OR\:\: IQ \gt 145 \\[3ex] $ Therefore 69 and 145 are the IQ scores that separate the unusual IQ scores from the usual IQ scores

$ (a.)\:\: Difference = 4939455 - 3670505 = \$1,268,950.00 \\[3ex] (b.)\:\: x = 4939455 \\[3ex] \mu = 3670505 \\[3ex] \sigma = 7775948 \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{4939455 - 3670505}{7775948} \\[5ex] z = \dfrac{1268950}{7775948} \\[5ex] z = 0.16 \\[3ex] The\:\:difference\:\:is\:\:0.16\:\:standard\:\:deviations \\[3ex] (c.)\:\: \underline{Obama's\:\:worth} \\[3ex] x = 3670505 \\[3ex] \mu = 4939455 \\[3ex] \sigma = 7775948 \\[3ex] z-score = \dfrac{3670505 - 4939455}{7775948} \\[5ex] z-score = -\dfrac{1268950}{7775948} \\[5ex] z-score = - 0.1631891057 \\[3ex] z-score \approx -0.16 \\[3ex] (d.)\:\:Because -2.00 \le -0.16 \le -2.00 \\[3ex] $ Obama's net worth is usual.

Sample size = total number of values = n = 36
Number of data values less than 65 = 27

$ Percentile\;\;of\;\;65 \\[3ex] = \dfrac{27}{36} * 100 \\[5ex] = \dfrac{2700}{36} \\[5ex] = 75th\;\;percentile \\[3ex] $ This means that 75% of the ages are lower than 65 years while 25% of the ages are higher than 65 years.
Alternatively, we can say that if the ages is divided into 100 parts; the age, 65 years separates the lowest 75% of the ages from the highest 25%.

$ Decile\;\;of\;\;65 \\[3ex] = \dfrac{27}{36} * 10 \\[5ex] = \dfrac{270}{36} \\[5ex] = 7.5th\;\;decile \\[3ex] $ If the ages dataset is divided in 10 parts, 65 years separates the lowest 7.5 of the ages from the highest 2.5.

$ Quintile\;\;of\;\;65 \\[3ex] = \dfrac{27}{36} * 5 \\[5ex] = \dfrac{135}{36} \\[5ex] = 3.75th\;\;quintile \\[3ex] $ If the ages dataset is divided in 5 parts, 65 years separates the lowest 3.75 of the ages from the highest 1.25.

$ Quartile\;\;of\;\;65 \\[3ex] = \dfrac{27}{36} * 4 \\[5ex] = \dfrac{108}{36} \\[5ex] = 3rd\;\;quartile \\[3ex] $ If the ages dataset is divided in 4 parts, 65 years separates the lowest 3 of the ages from the highest 1.

$ n = 36 \\[3ex] xth\;\;position = \dfrac{75}{100} * 36 = \dfrac{2700}{100} = 27 \\[5ex] xth\;\;position = \dfrac{7.5}{10} * 36 = \dfrac{270}{10} = 27 \\[5ex] xth\;\;position = \dfrac{3.75}{5} * 36 = \dfrac{135}{5} = 27 \\[5ex] xth\;\;position = \dfrac{3}{4} * 36 = \dfrac{108}{4} = 27 \\[5ex] \implies \\[3ex] xth\;\;position \\[3ex] = \dfrac{27th\;\;position + 28th\;\;position}{2} \\[5ex] = \dfrac{63 + 65}{2} \\[5ex] = \dfrac{128}{2} \\[5ex] = 64 \\[3ex] $ 75th percentile = 7.5th decile = 3.75th quintile = 3rd quartile = 64

Student: Mr. C, are we not supposed to get 65 as the answer?
Teacher: Yes. But we got 64. This is the correct answer.
I want you to see it this way: 64 is the boundary for the 3rd quartile (75th percentile)
65 is included in that part. But 64 is the boundary.
Be it as it may: please
NOTE: Some textbooks/online resources calculate these formulas (formulas that deal with converting a quantile to a data value) differently.
They may have it this way:

$ xth\:\:position = \dfrac{yth\:\:Percentile}{100} * (total\:\:number\:\:of\:\:values + 1) \\[5ex] $ If we use this formula, we will get the answers we got in Question Number (5.) (which we makes sense).
However, if we decide to calculate the median of that data with this formula, it will be different from the median value calculated with the formulas in the link above (main formulas).
There is still some confusion that the main formulas should only apply to the median.
Be it as it may, please use the main formulas (which is provided for you on the tests/exams)

(a.) 392.8 million BTU = Q₃ = 75%
More than 75% = 100% = 75% = 25%
25% of the states consumed more than 392.8 million BTUs per capita.

(b.) 242.9 million BTUs = Q₁ = 25%
More than 25% = 100% - 25% = 75%
75% of the states consumed more than 242.9 million BTUs per capita.

(c.) 304.7 million BTUs = Q₂ = Median = 50%
More than 50% = 100% - 50% = 50%
50% of the states consumed more than 304.7 million BTUs per capita.

$ (d.) \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 392.8 - 242.9 = 149.9\;million\;BTUs \\[3ex] $ (e.) Interpretation of IQR: The middle​ 50% of the states have values of total energy consumption per capita that vary by as much 149.9 million BTUs.

$ \mu = 78 \\[3ex] \sigma = 6 \\[3ex] z = 2 \\[3ex] x = ? \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] x - \mu = z * \sigma \\[3ex] x = z * \sigma + \mu \\[3ex] x = 2 * 6 + 78 \\[3ex] x = 12 + 78 \\[3ex] x = 90 $

(a.) The sport with the most expensive ticket prices is Sport C because it has the greatest maximum ticket price.
(b.) The sport with the most expensive ticket prices is Sport D because it has the least minimum ticket price.
(c.) A typical ticket for Sport A is less expensive than a typical ticket for Sport ​B, since Sport A has a lower median than Sport B.
Sport B has more variability in ticket​ prices, as shown by a higher IQR (the length of the box...without the whiskers).
Sport A has potential outliers representing unusually high-priced tickets.
Sport B has potential outliers representing unusually high-priced tickets.

The data is already arranged in ascending order.

$ n = 15 \\[3ex] Min = 70 \\[3ex] xth\;\;position = \dfrac{1}{4} * 15 = 3.75 \approx 4th\;\;position = 71 \\[5ex] Q_1 = 71 \\[3ex] xth\;\;position = \dfrac{2}{4} * 15 = 7.5 \approx 8th\;\;position = 74 \\[5ex] Q_2 = 74 \\[3ex] xth\;\;position = \dfrac{3}{4} * 15 = 11.25 \approx 12th\;\;position = 78 \\[5ex] Q_3 = 78 \\[3ex] Max = 79 \\[3ex] $ The five-number summary: 70, 71, 74, 78, 79

(a.) The sketch of the boxplot is:



$ (b.) \\[3ex] Q_1 = 8.009 \\[3ex] Q_3 = 33.313 \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 33.313 - 8.009 \\[3ex] IQR = 25.304 \\[3ex] LF = Q_1 - 1.5IQR \\[3ex] LF = 8.009 - 1.5(25.304) \\[3ex] LF = 8.009 - 37.956 \\[3ex] LF = -29.947 \\[5ex] UF = Q_3 + 1.5IQR \\[3ex] UF = 33.313 + 37.956 \\[3ex] UF = 71.269 \\[3ex] $ The length of the whiskers is determined as follows:
The minimum value is greater than or equal to the left limit (lower fence) of −29.947, so the left whisker is drawn to the minimum.
The maximum value is greater than the right limit (upper fence) of 71.269, so the right whisker is drawn to the largest non-outlier.

The data is in ascending order.

$ n = 18 \\[3ex] Min = 5 \\[3ex] xth\;\;position = \dfrac{1}{4} * 18 = 4.5 \approx 5th\;\;position = 10 \\[5ex] Q_1 = 10 \\[3ex] xth\;\;position = \dfrac{2}{4} * 18 = 9 = \dfrac{9th\;\;position + 10th\;\;position}{2} = \dfrac{12 + 17}{2} = 14.5 \\[5ex] Q_2 = 14.5 \\[3ex] xth\;\;position = \dfrac{3}{4} * 18 = 13.5 \approx 14th\;\;position = 37 \\[5ex] Q_3 = 37 \\[3ex] Max = 64 \\[3ex] $ The five-number summary is: 5, 10, 14.5, 37, 64

First, let us list the five-number summary
Then, we calculate the lower fence and upper fence
Then, we make our decision based on the comparing the minimum and the lower fence and/or the maximum and upper fence.

$ (a.) \\[3ex] 44, 78, 80, 87, 95 \\[3ex] minimum = 44 \\[3ex] Q_1 = 78 \\[3ex] Q_2 = 80 \\[3ex] Q_3 = 87 \\[3ex] maximum = 95 \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 87 - 78 = 9 \\[3ex] 1.5IQR = 1.5(9) = 13.5 \\[3ex] LF = Q_1 - 1.5IQR \\[3ex] LF = 78 - 13.5 = 64.5 \\[3ex] 44 \lt 64.5 \\[3ex] $ The minimum value is less than the lower fence.
Between the minimum value and the lower fence, there may be some values greater than the minimum but less than the lower fence.
These values are unknown outliers.
Hence, the boxplot cannot be drawn for this dataset because the boxplot must mark all potential outliers.
Since the minimum is lower than the left​ limit, there may be other unknown outliers between the minimum and left​ limit, and so the boxplot cannot mark all potential outliers.

$ (b.) \\[3ex] 51, 72, 80, 86, 100 \\[3ex] minimum = 51 \\[3ex] Q_1 = 72 \\[3ex] Q_2 = 80 \\[3ex] Q_3 = 86 \\[3ex] maximum = 100 \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 86 - 72 = 14 \\[3ex] 1.5IQR = 1.5(14) = 21 \\[3ex] LF = Q_1 - 1.5IQR \\[3ex] LF = 72 - 21 = 51 \\[3ex] 51 = 51 \\[3ex] $ The minimum is the lower fence.
So, there are no unknown outliers based on the lower fence.
Let us compare the maximum with the upper fence.

$ UF = Q_3 + 1.5IQR \\[3ex] UF = 86 + 21 = 107 \\[3ex] $ The maximum is lower than the upper fence.
This is in order because there is no other value in the dataset greater than the maximum.
It is possible to draw a boxplot based on this information.
Both the minimum and maximum are within the bounds of the left limit and right​ limit, which means that all potential outliers can be displayed.
This is necessary to construct the boxplot.

(a.) 43.7 million BTUs = Q₁ = 25%
25% of the states consumed fewer than 43.7 million BTUs per capita.

(b.) 140.6 million BTUs = Q₃ = 75%
75% of the states consumed fewer than 140.6 million BTUs per capita.

(c.) Median = 94.9 million BTU = 50%
More than 50% = 100 - 50% = 50%
50% of the states consumed more than 94.9 million BTUs per capita.

$ (d.) \\[3ex] \underline{Industrial\;\;BTU} \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 140.6 - 43.7 = 96.9\;million\;BTUs \\[5ex] \underline{Total\;\;BTU} \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 380.4 - 228.4 = 152\;million\;BTUs \\[5ex] 152\;million\;BTUs \gt 96.9\;million\;BTUs \\[3ex] $ There is more variability in Total Energy Consumption because it has an interquartile range of 152 million BTUs versus an interquartile range for Industrial Energy Consumption of 96.9 million BTUs.

Scale:

$ \dfrac{12 - 8}{2} = \dfrac{4}{2} = 2 \\[5ex] 8, 10, 12, 14, 16, 18, 20 \\[3ex] $ (a.) Minimum Poverty Rates:
A = about 10.2
B = about 8.4
C = about 11.8
D = about 8.2
The regions from highest to lowest median poverty rate are: C, A, B, D

(b.) Interquartile Range is the length of the box: (approximate lengths are calculated)

$ IQR = Q_3 - Q_1 \\[3ex] A = 13 - 11 = 2 \\[3ex] B = 12.8 - 8.4 = 4.4 \\[3ex] C = 18.2 - 14.4 = 3.8 \\[3ex] D = 13.7 - 10.4 = 3.3 \\[3ex] $ The regions from lowest to highest interquartile range are: A, D, C, B

(c.) An unusually low or an unusually high poverty​ rate is an outlier.
Region A has a state with an unusually low poverty rate.
Region B has no state with an unusually low or high poverty rate.
Region C has a state with an unusually low poverty rate.
Region D has no state with an unusually low or high poverty rate.
Regions that have states with unusually low poverty rates have dots on the left side of their boxplots.
Regions that have states with unusually high poverty rates have dots on the right side of their boxplots.

(d.) Region A has the least amount of variability because it has the lowest interquartile range.

(e.) The range depends on only two observations, while the interquartile range depends on many observations and is therefore more reliable.

$ n = 500 \\[1em] \underline{65\:years} \\[1em] 75th\:\:percentile = \dfrac{75}{100} * 500 = 75 * 5 = 375 \\[2em] \underline{15\:years} \\[1em] 25th\:\:percentile = \dfrac{25}{100} * 500 = 25 * 5 = 125 \\[2em] \underline{Between\:15\:years\:\:and\:\:65\:years} \\[1em] 375 - 125 = 250\:people $

(a.) Some of these data have outliers​ and/or are​ skewed, and the median and interquartile range are resistant to outliers.

(b.) Scale on the x-axis:

$ \dfrac{50 - 0}{2} = \dfrac{50}{2} = 25 \\[5ex] $ The median for Region B (between 150 and 175) is about 170 people per square mile.
The median for Region C (between 500 and 525) is about 520 people per square mile.
The median for Region D (between 75 and 100) is about 95 people per square mile.

(c.) Interquartile Range is the length of the box: (approximate lengths are calculated)

$ IQR = Q_3 - Q_1 \\[3ex] A = 13 - 11 = 2 \\[3ex] B = 12.8 - 8.4 = 4.4 \\[3ex] C = 18.2 - 14.4 = 3.8 \\[3ex] D = 13.7 - 10.4 = 3.3 \\[3ex] $ The regions from lowest to highest interquartile range are: A, D, C, B

(a.) The median BA percentage for Region A is approximately is 23.6
(b.) The median BA percentage for Region B is approximately is 25.4
(c.) A greater percentage of the population has a BA degree in Region B because the median is larger for Region B.
(d.) Region B has greater variation than Region A because the interquartile range (IQR) for Region A is less than that for Region B.

Fences: Lower Fence and Upper Fence determine outliers of data.
A data value is an outlier if it is less than the lower fence or greater than the upper fence.
Because the question asked for the height of the shortest player, we shall be concerned with only the Lower Fence.
Data is already sorted

$ n = 10 \\[3ex] \dfrac{1}{4} * 10 \\[5ex] = 2.5th \approx 3rd\;\;position \\[3ex] Q_1 = 185 \\[5ex] \dfrac{3}{4} * 10 \\[5ex] = 7.5th \approx 8th\;\;position \\[3ex] Q_3 = 194 \\[5ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 194 - 185 \\[3ex] IQR = 9 \\[3ex] LF = Q_1 - 1.5(IQR) \\[3ex] LF = 185 - 1.5(9) \\[3ex] LF = 185 - 13.5 \\[3ex] LF = 171.5 \\[3ex] $ Because the height of the shortest player is 170 cm and 170 < 171.5; the height of the shortest player is an outlier.

(a.) The sketch of the boxplot is:



$ (b.) \\[3ex] Q_1 = 78 \\[3ex] Q_3 = 91 \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 91 - 78 \\[3ex] IQR = 13 \\[3ex] LF = Q_1 - 1.5IQR \\[3ex] LF = 78 - 1.5(13) \\[3ex] LF = 78 - 19.5 \\[3ex] LF = 58.5 \\[5ex] UF = Q_3 + 1.5IQR \\[3ex] UF = 91 + 19.5 \\[3ex] UF = 110.5 \\[3ex] $ The whiskers extend to the most extreme values that are not potential outliers.
The left whisker extends to the smallest observation that is greater than or equal to 58.5​, and the right whisker extends to the largest observation that is less than or equal to 110.5