Step 1
Data Set A- 7,7,7,9,9,9,10
Data Set B- 4,6,6,6,8,9,9,9,10,10,10
Step 1
Step 2
Data Set A
x |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
y |
7 |
7 |
7 |
9 |
9 |
9 |
10 |
Data Set B
x |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
y |
4 |
6 |
6 |
6 |
8 |
9 |
9 |
9 |
10 |
10 |
10 |
Step 2
A= c(7,7,7,9,9,9,10)
B= c(4,6,6,6,8,9,9,9,10,10,10)
mean(A)=sum of the observations/7 = 8.285714
standard deviation(A)=1.253566
median(A)= 9 (the middle most observation)
mode= 3 median - 2 mean=10.42857=10 (approx)
Q1(A)=7 (25% th i.e. the 2 nd observation)
Q3(A)=9 (75%th i.e the 6th observation)
QD(A)={Q3(A)-Q1(A)}/2=(9-2)/2=3.5
adj factor =QD(A).(1.5)=5.25
inner fences(A) are {Q3(A)+5.25, Q1(A)-5.25} i.e. {14.25,1.75}
i.e. the boundaries of our outlier fences for data set A is (1.75, 14.25)
Now Data set A= c(7,7,7,9,9,9,10)
Clearly no data of data set A lies outside the boundary .
similarly,
mean(B)=7.909091
standard deviation(B)= 2.071451
median(B)= 9
mode= 3 median - 2 mean=11.18182=11 (approx)
Q1(B)=6
Q3(B)=10
QD(B)={Q3(B)-Q1(B)}/2=(10-6)/2 =2
adj factor (B)=QD(B).(1.5)=3
inner fences(B) are {Q3(B)+3, Q1(B)-3} i.e. {13,3}
i.e. the boundaries of our outlier fences for data set B is (3,13)
B= c(4,6,6,6,8,9,9,9,10,10,10)
Clearly, In data set B no data lies outside the boundary (3, 13).
So clearly not a single data is found neither in data set A nor in the data set B. So all the mean, sd, median, quantiles will be same as we calculate already above.
Please come again with the other questions. Thank you
Step 1 Data Set A- 7,7,7,9,9,9,10 Data Set B- 4,6,6,6,8,9,9,9,10,10,10 Step 1 Find the sample mean,...
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