What percentage of the general U.S. population are high-school dropouts? The Statistical Abstract of the United States, 120 th Edition, gives the percentage of high-school dropouts by state. For convenience, the data are sorted in increasing order. $$ \begin{array}{c} 5,6,7,7,7,7,8,8,8,8,8,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,11,11,11,11,11,11 \ 11,11,12,12,12,12,13,13,13,13,13,13,14,14,14,14,14,15 \end{array} $$ (a) Which percentile is the value 10 ? (Round to the nearest whole value above.) 10 is the $$ \square $$ th percentile. (b) Find the 95th percentile on the data. Locator $$ = $$ $$ \square $$ $$ 95 \mathrm{th} \text { percentile }=14 $$ (c) Find the $$ Q_{1} $$ value on the data. $$ \text { Locator }=13 $$ $$ Q_{1}=9 $$ (d) Find the Q3 value on the data. $$ \text { Locator }=\square $$ $$ Q_{3}=12 $$ (e) Find the $$ Q_{2} $$ value on the data. $$ Q_{2}=10 $$ (f) Find the $$ I Q R $$ of the data. $$ I Q R=3 $$ (g) What are the Upper and Lower limits of the boxplot? $$ \text { Upper Limit }=16.5 $$ Lower Limit $$ =4.5 $$ (h) With five-number-summary (in the order Min = $$ Q_{1}=Q_{2}-Q_{3}-\operatorname{Max} $$ ) being 5-9-10=12 15 , are there any outliers? The only outlier is 5 The only outlier is 15 Outliers are 15, and 14 Outliers are 5, and 15. o No outliers

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### Given Data
The data set of high-school dropout percentages is:
$$
5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15
$$

### (a) Which percentile is the value 10?
To find the percentile of a value in a data set, we use the formula:
$$
P = \left( \frac{N_L}{N} \right) \times 100
$$
where $$ N_L $$ is the number of values less than the value of interest, and $$ N $$ is the total number of values.

1. Count the number of values less than 10:
   - Values less than 10: $$ 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9 $$ (total of 18 values)
   
2. Total number of values $$ N = 46 $$.

Now, calculate the percentile:
$$
P = \left( \frac{18}{46} \right) \times 100 \approx 39.13
$$
Rounding up gives us the 40th percentile.

**Answer:** 10 is the $$ 40 $$th percentile.

### (b) Find the 95th percentile on the data.
To find the 95th percentile, we calculate the locator:
$$
L = \frac{95}{100} \times (N + 1) = 0.95 \times 47 = 44.65
$$
Rounding up gives us the 45th position. The value at the 45th position is 14.

**Answer:** $$ 95 \text{th percentile } = 14 $$

### (c) Find the $$ Q_{1} $$ value on the data.
To find $$ Q_1 $$:
$$
L = \frac{25}{100} \times (N + 1) = 0.25 \times 47 = 11.75
$$
Rounding up gives us the 12th position. The value at the 12th position is 9.

**Answer:** $$ Q_{1} = 9 $$

### (d) Find the $$ Q_{3} $$ value on the data.
To find $$ Q_3 $$:
$$
L = \frac{75}{100} \times (N + 1) = 0.75 \times 47 = 35.25
$$
Rounding up gives us the 36th position. The value at the 36th position is 12.

**Answer:** $$ Q_{3} = 12 $$

### (e) Find the $$ Q_{2} $$ value on the data.
The median $$ Q_2 $$ is found at:
$$
L = \frac{50}{100} \times (N + 1) = 0.5 \times 47 = 23.5
$$
Rounding up gives us the 24th position. The value at the 24th position is 10.

**Answer:** $$ Q_{2} = 10 $$

### (f) Find the $$ IQR $$ of the data.
The interquartile range $$ IQR $$ is calculated as:
$$
IQR = Q_3 - Q_1 = 12 - 9 = 3
$$

**Answer:** $$ IQR = 3 $$

### (g) What are the Upper and Lower limits of the boxplot?
- **Upper Limit**: $$ Q_3 + 1.5 \times IQR = 12 + 1.5 \times 3 = 12 + 4.5 = 16.5 $$
- **Lower Limit**: $$ Q_1 - 1.5 \times IQR = 9 - 1.5 \times 3 = 9 - 4.5 = 4.5 $$

**Answer:** Upper Limit = 16.5, Lower Limit = 4.5

### (h) Are there any outliers?
To find outliers, we check for values outside the limits:
- Outliers are values less than 4.5 or greater than 16.5.
- The data set has no values below 5 or above 15.

**Answer:** No outliers.

### Summary of Answers
(a) 40  
(b) 14  
(c) 9  
(d) 12  
(e) 10  
(f) 3  
(g) Upper Limit = 16.5, Lower Limit = 4.5  
(h) No outliers.
(a) 40 (b) 14 (c) 9 (d) 12 (e) 10 (f) 3 (g) Upper Limit = 16.5, Lower Limit = 4.5 (h) No outliers

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