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KEY DAY 7 ACTIVITY - HOW DID I DO (PERCENTILES^LJ Z-SCORES AND LINEAR TRANSFORMATIONS)

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Day 7 Activity “How Did I Do?” (Percentile, z-Scores & Linear Transformation)
AP Statistics
Name _________________________________________ Block _________ Date _______________
How well did you do on the Chapter 1 Test? How well did you do relative to your classmates?
Here are the results of a random sample of 20 of the Chapter 1 Tests, along with a dotplot and summary statistics.
Test Scores
61 65 65 73 75 77 78 78 79 80 80 80 80 81 81 88 89 93 98 99
1. Biff scored a 65. What is Biff’s percentile?
Note: Percentile is the percent of values strictly LESS or EQUAL TO.
3/20=0.15 or 15th percentile
2. Given that the average test score was 80, and standard deviation was 10, was Biff above or below the mean? By
how many points? By how many standard deviations?
Biff’s score was below the mean. Below by 15 points. 𝑧 =
65−80
10
= −1.5, 1.5 standard deviation below the mean.
3. Marty scored an 88. What is Marty’s percentile?
16/20=0.80 or 80th percentile
4. Was Marty above or below the mean? By how many points? By how many standard deviations?
Marty’s score was above the mean. Above by 8 points. 𝑧 =
88−80
10
= 0.8, 0.8 standard deviation above the mean.
A z-score is defined as the number of standard deviations above or below the mean.
5. Write a formula for calculating a z-score.
𝑧=
𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
6. Goldie scored a 98 on the Chapter 1 Test. Find and interpret the z-score. 𝑧 =
98−80
10
= 1.8
Goldie’s Chapter 1 test score is 1.8 standard deviation above the mean.
Linear Transformation
Number of Pairs of Shoes
14
7
13
7
4
10
12
8
12
7
3
3
10
4
8
2
12
14
6
9
5
4
4
4
Mean 𝑋
St. Dev. 𝑆𝑋
Variance 𝑆𝑋 2
Median (𝑀𝑒𝑑)
IQR
Range
7.583
3.80
14.428
7
7
12
Add 7 to each Element
Mean 𝑋
St. Dev. 𝑆𝑋
Variance 𝑆𝑋 2
Median (𝑀𝑒𝑑)
IQR
Range
14.583
3.72
14.428
14
7
12
Adding (or subtracting) a constant to each value in a set of data changes the center (mean & Median) but does not the
shape or the spread (SD, IQR, Range) of a distribution. Adding or subtracting slides the entire distribution a distance
equivalent to the constant.
Multiply Each Element of the Original Data by four.
Mean 𝑋
St. Dev. 𝑆𝑋
Variance 𝑆𝑋 2
Median (𝑀𝑒𝑑)
IQR
Range
30.333
15.193
230.827
28
28
48
Multiplying (or dividing) each value in a set of data by a positive constant does not change the basic shape of the
distribution. Both the center (mean and median) and the spread (SD, IQR, Range) are multiplied or divided by that
number. This transformation amounts to a rescaling of the distribution.
Multiply Each Element of the Original Data by four, then add seven.
Mean 𝑋
St. Dev. 𝑆𝑋
Variance 𝑆𝑋 2
Median (𝑀𝑒𝑑)
IQR
Range
37.333
15.193
230.827
35
28
48
The Influence of Outliers
A summary statistic is resistant if the summary statistic is not changed very much when an outlier is removed from the set of
data. If the summary statistic tends to be affected by the removal of outliers, it is sensitive to outliers.
Check Your Understanding:
1.Knoebels Amusement Park in Elysburg, Pennsylvania, has earned acclaim for being an affordable, family-friendly
entertainment venue. Knoebels does not charge for general admission or parking, but it does charge customers for
each ride they take. How much do the rides cost at Knoebels? The figure shows a dot- plot of the cost for each of
22 rides in a recent year, along with summary statistics.
a.
Suppose you convert the cost of the rides from dollars to cents ($1=100 cents). Describe the shape, mean,
and standard deviation of the distribution of ride cost in cents.
The shape of the distribution of the cost for each of 22 rides remains the same, skewed to the right.
The center (mean) is multiplied by 100 cents: 1.705 ×100=170.5 cents.
The standard deviation changes to: 0.447 ×100=44.7 cents.
b. Knoebels’ managers decide to increase the cost of each ride by 25 cents. How would this the shape, center,
and variability of this distribution compare with the distribution of cost in Question 1?
The shape of the distribution of the cost for each of 22 rides remains the same, skewed to the right.
The center (mean) is increased by 25 cents: 170.5+25=195.5 cents.
The standard deviation stays the same at 44.7 cents.
c. Now suppose you convert the increased costs from Question 2 to z-scores. What would be the shape, mean,
and standard deviation of this distribution? Explain your answers.
The shape of the distribution of the cost for each of 22 rides remains the same, skewed to the right.
The center (mean) is shifted down to 0 cents.
The standard deviation becomes 1 cent.
2. Mrs. Munson is concerned about how her daughter’s height and weight compare with those of other girls of the
same age. She uses an online calculator to determine that her daughter is at the 87th percentile for weight and
the 67th percentile for height. Explain to Mrs. Munson what these values mean.
87% of girls of the same age have a weight that is less than or equal to her daughter’s weight. 67% of other girls
of the same age have a height that is less than or equal to her daughter’s height.
a.
The graph displays the cumulative relative frequency of the lengths of phone calls made from the
mathematics department office at Gabalot High last month.
b. About what percent of calls lasted less than 30 minutes? 30 minutes or more?
About 65% are less than 30 minutes.
About 35% are 30 minutes or more.
c.
Estimate Q1, Q3, and the IQR of the distribution of phone call length.
Q1≈12 min.
Q3≈32 min.
IQR= Q3-Q1=20 min.
3.The following data set represents Mr. T’s students’ scores on the final.
a. If you got a 90 on the final, what percentile does that place you in?
66
69
71
71 72 73.5 74 74 76.5
76.5 77.5
78
81 81
85
86 86
86
87 87.5 88.5 89 90
92
92 100
23/26= 88th percentile
b. If you are in the 46 percentile, what did you score?
46%(26) = 11.96 or 12th spot= 78
th
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