CHEGG The accompanying observations are precipitation values during March over a 30-year period in Minneapolis-St. Paul. 0.32, 0.47, 0.52, 0.59, 0.77, 0.81, 0.81, 0.90, 0.96, 1.18, 1.20, 1.20, 1.31, 1.35, 1.43, 1.51, 1.62, 1.74, 1.87, 1.89, 1.95, 2.05, 2.10, 2.20, 2.48, 2.81, 3.00, 3.09, 3.37, 4.75 (a) Construct a normal probability plot for this data set. Hint: You can use the R software. Issue this command (cut and paste the 2 rows of code below in the R command line): qqnorm(c(0.32,0.47,0.52,0.59,0.77,0 .81,0.81,0.90,0.96,1.18,1.20,1.20,1 .31,1.35,1.43,1.51,1.62,1.74,1.87,1 .89,1.95, 2.05,2.10,2.20,2.48,2.81,3.00,3.09, 3.37,4.75))

Mean 1.5806666671.769333333

Standard Deviation 0.8907417991.123161779

Step-by-step explanation:

When we run the commands in the R software we get the graphs as follows.

The commands were

qqnorm(c(0.32,0.52,0.77,0.81,0.96,1.2,1.31,1.43,1.62,1.87,1.95,2.1,2.48,3,3.37))

and

qqnorm(c(0.47,0.59,0.81,0.9,1.18,1.2,1.35,1.51,1.74,1.89,2.05,2.2,2.81,3.09,4.75))

We can plot the graphs in any other plotting software .

The results would be same .

We see there's a slight difference between the two graphs.

One curve goes above ( Minneapolis) for  0.4 and the other ( for St Paul) is below 0.4

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

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step-by-step explanation: