Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot. Look at the QQ plot and say whether the data are plausibly from a normalpopulation or whether they are evidence against a normal population. (Hint: In each case, to get somecontext for your decision, make a few QQ plots from samples of the same size that did come from a normalpopulation. Please do not include these context plots in your submission.)(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8, 10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10 .7,11,9.4)(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9, 1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)(c)da ta = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,1 2,14.4,12.8,11.2,10.9,11.7,11.7,11. 6)

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: