On this assignment, round all answers (when necessary) to exactly three decimal places. The numbers of construction workers for various projects are 32 20 25 52 16 21 28 35 23 41 46 17 23 27 The mean of this data is The standard deviation of this data is The five-number summary is { , , , , } The Interquartile Range (IQR) is Are there any outliers according to the IQR rule (type 0 for yes, 1 for no) Replacing the largest number with results in the smallest whole number outlier.

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ChiKesselman ChiKesselman · Answer 1 · 2019-09-26T08:24:20+02:00

Answer:

Mean = 29

S.D = 10.95

NO outlier in the data

Step-by-step explanation:

We are given the following data:

n = 14

Construction Workers: 32, 20, 25, 52, 16, 21, 28, 35, 23, 41, 46, 17, 23, 27

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

Mean = [tex]\frac{406}{14} = 29[/tex]

Standard Deviation =

[tex]\sqrt{\frac{9,+ 81+16+ 529+ 169+ 64+ 1+36+ 36+ 144+ 289+ 144+ 36+ 4}{13}}\\= \sqrt{\frac{1558}{13} } = 10.94[/tex]

Five number summary:

Data = 16,17,20,21,23,23,25,27,28,32,35,41,46,52

Minimum = 16

Maximum = 52

Median = Mean of 25 and 27 = 26

First Quartile = 21

Third Quartile = 35

Interquartile range = [tex]Q_3 - Q_1[/tex] = 35 - 21 = 14

Outliers:

[tex]\text{Lower limit} = Q_1 - (1.5)IQR = 21 - 21 = 0\\\text{Upper limit} = Q_3 + (1.5)IQR = 35 + 21 = 56\\[/tex]

There is no outlier in the data.