The following describes a sample. The information given includes the five number summary, the sample size, and the largest and smallest data values in the tails of the distribution.

Five number summary: (160,210,220,250,270); n=500
Tails: 160,165,167,171,175,...,268,269,269,269,270,270

Clearly identify any outliers using the IQR method. Select all that apply.

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

[tex] IQR = Q_3 -Q_1 = 250-210 =40[/tex]

For this case we need to find the lower and upper limit with the following formulas:

[tex] L= Q_1 - 1.5*IQR = 210 -1.5*40 =150[/tex]

[tex] U= Q_3 + 1.5*IQR = 250 +1.5*40 =310[/tex]

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

topeadeniran2 topeadeniran2 · Answer 2 · 2022-02-03T14:32:53+01:00

From the information in the sample, the five number summary include 160, 210, 220, 250, and 270.

Solving the sample

From the information given, the lower fence will be:

= 210 - (1.5 × 40) = 150

The upper fence will be:

= 250 + (1.5 × 40) = 310

From the information given, the number of values is 500. The five number summary include 160, 210, 220, 250, and 270.

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