Use five classes. 2. Using the charted information shown below, make a stem and

Question

Use five classes. 2. Using the charted information shown below, make a stem and leaf display for each of the following respective headings. Tar, Nicotine, and Co. Are there any outliners in each? (See the definition on page 73.) Nicotine Tar 0.78 0.74 0.13 1.26 1.08 10.2 Benson & Hedges Bull Durham Camel Lights Carlton Chesterfield Golden Lights 16.0 29.8 8.0 4.1 15.0 3.6 16.6 23.5 10.2 5.4 15.0 9.0 12.3 16.3 11.4 9.0 1.0 17.0 2.8 15.8 1.06 Old Gold Pall Mall Lights 18.5 2.6 17.5 4.9 5.9 8.5 0.6 13.9 14.9 0.40 1.04 0.76 0.42 1.01 0.61 Salem Ultra 12.4 16.6 4.9 13.7 15.1 7.8 14.5 7.3 1.02 1.01 0.90 Viceroy Rich Light Virginia Slims Winston Lights Lark Lights 5.2 12.0 1.02 13.0 14.4 10.0

Explanation / Answer

## By using Minitab

Stem-and-Leaf Display: Tar

Stem-and-leaf of Tar   N = 25

Leaf Unit = 1

Here Q1= 8.3; Q3=15.15

hence IQR=15.15-8.3 =6.85

therefore

Q1-1.5*IQR>Outliers or Q3+1.5*IQR < Outliers

8.3 -1.5*6.85 =-1.975 or 15.15 +1.5*6.85 = 25.425

Since 29.8 >25.425

From The above stem and leaf the leaf 29.8 is the outliers in the data.

Stem-and-Leaf Display: Nicotine

Stem-and-leaf of Nicotine   N = 25

Leaf Unit = 0.1

Here Q1 =0.68 Q3 = 1.03 IQR=0.35

Q1-1.5*IQR =0.68-1.5*0.35 or Q3+1.5*IQR =1.03+1.5*0.35   

=0.155 =1.555

Since 0.13<0.155 and 2.3>1.555

From the above stem and leaf the leaf 0.13 and 2.3 are the outliers in the data.

Stem-and-Leaf Display: CO

Stem-and-leaf of CO   N = 25

Leaf Unit = 1

Here Q1=9.75 Q3=15.65 IQR= 5.9

hence

Q1-1.5*IQR = 9.75 -1.5*5.9 =0.9

Q3+1.5*IQR = 15.65 +1.5*5.9 =24.5

Since no point lies out side (Q1-1.5*IQR,Q3+1.5*IQR)

There are no outliers in this data

3 0 144 9 0 778889 (8) 1 12223444 8 1 5555667 1 2 1 2 9

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