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Lesson Example Discussion Quiz: Class Homework

Step-5

Title: Box and Whisker Plots, Identifying Outliers

Grade: 1300-a Lesson: S4-L6

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step	Type	Explanation	Answer
1	Problem	Identify any outlier(s) of the data set. 76, 72, 64, 93, 80, 78, 96, 75, 70, and 72.
2	Step	Order the data in ascending order: 64, 70, 72, 72, 75, 76, 78, 80, 93, and 96.
3	Step	First, we have to find the first quartile and third quartile ranges Q1 and Q3. → Q1: The median of the lower half (64, 70, 72, 72, and 75) Q1 = 72. → Q3: The median of the upper half (76, 78, 80, 93, and 96) Q3 = 80.
4	Step	Calculate the Interquartile Range (IQR): IQR = 80 - 72 = 8.
5	Step	Identify potential outliers → \$"Lower bound" = "Q"1 - 1.5 \times "IQR"\$ \$ = 72 - 1.5 \times 8\$ \$= 60\$. → \$"Upper bound" = "Q"3 + 1.5 \times "IQR"\$ \$= 80 - 1.5 \times 8\$ \$= 92\$.
6	Step	Identify Outliers Any value below 60 or above 92 is an outlier. → The dataset is: 76, 72, 64, 93, 80, 78, 96, 75, 70, and 72. → The value 93 is above 92. → The value 96 is above 92.
7	Step	Therefore, the outliers in this dataset are 93 and 96.
8	Choice.A	This option is incorrect because 96 is an outlier, but 94 is not an outlier it is within the upper bound	94, 96
9	Choice.B	This option is incorrect because only 93 is questionable based on the IQR, and 92 is even lower than the data set	93, 92
10	Choice.C	This option is incorrect because 95 is closer to the upper bound than 93, and 98 is a clear outlier beyond the IQR range	95, 98
11	Choice.D	This option is correct because it is the most accurate answer. Both 93 and 96 fall outside the typical range of the data, based on IQR	93, 96
12	Answer	Option	D
13	Sumup	Can you summarize what you’ve understood in the above steps?

Step

Type

Explanation

Answer

Problem

Identify any outlier(s) of the data set.
76, 72, 64, 93, 80, 78, 96, 75, 70, and 72.

Step

Order the data in ascending order: 64, 70, 72, 72, 75, 76, 78, 80, 93, and 96.

Step

First, we have to find the first quartile and third quartile ranges Q1 and Q3.
→ Q1: The median of the lower half (64, 70, 72, 72, and 75)
Q1 = 72.

→ Q3: The median of the upper half (76, 78, 80, 93, and 96)
Q3 = 80.

Step

Calculate the Interquartile Range (IQR): IQR = 80 - 72 = 8.

Step

Identify potential outliers
→ \$"Lower bound" = "Q"1 - 1.5 \times "IQR"\$
\$ = 72 - 1.5 \times 8\$
\$= 60\$.
→ \$"Upper bound" = "Q"3 + 1.5 \times "IQR"\$
\$= 80 - 1.5 \times 8\$
\$= 92\$.

Step

Identify Outliers
Any value below 60 or above 92 is an outlier.
→ The dataset is: 76, 72, 64, 93, 80, 78, 96, 75, 70, and 72.
→ The value 93 is above 92.
→ The value 96 is above 92.

Step

Therefore, the outliers in this dataset are 93 and 96.

Choice.A

This option is incorrect because 96 is an outlier, but 94 is not an outlier it is within the upper bound

94, 96

Choice.B

This option is incorrect because only 93 is questionable based on the IQR, and 92 is even lower than the data set

93, 92

Choice.C

This option is incorrect because 95 is closer to the upper bound than 93, and 98 is a clear outlier beyond the IQR range

95, 98

Choice.D

This option is correct because it is the most accurate answer. Both 93 and 96 fall outside the typical range of the data, based on IQR

93, 96

Answer

Option

Sumup

Can you summarize what you’ve understood in the above steps?