Question

4. 150 Weights Mean: 168 pounds Standard Deviation: 35 pounds

Based on the Range Rule of Thumb, what is the minimum ≈ ______ and maximum≈ ______? How do these compare to the actual minimum and maximum in the data set below?

Based on the Range Rule of Thumb, what is the Estimated Standard Deviation: _______

How close is actual standard deviation of 35 to the one estimated using the range rule of thumb?

100 110 115 120 120 120 120 120 120 120 125 125 125 130 130 130

130 130 130 130 130 130 135 135 135 140 140 140 140 140 140 140

140 140 140 140 140 140 145 145 145 145 145 150 150 150 150 150

150 150 150 150 150 150 155 155 160 160 160 160 160 160 160 160

160 160 160 160 160 160 160 160 161 165 165 165 165 170 170 170

170 170 170 170 170 170 170 170 175 175 175 175 180 180 180 180

180 180 180 180 180 180 180 180 180 180 180 180 180 185 185 185

185 185 190 190 190 190 190 190 190 190 200 200 200 200 200 200

200 200 200 200 200 200 209 210 210 210 215 215 220 220 225 230

240 250 257 260 270 350

Question 5

5. 100 Heights Mean 64.3 in Standard Deviation 2.8 in

Based on the Range Rule of Thumb, what is the minimum ≈ ______ and maximum≈ ______? How do these compare to the actual minimum and maximum in the data set below?

Based on the Range Rule of Thumb, what is the Estimated Standard Deviation: _______

How close is actual standard deviation of 2.8 to the one estimated using the range rule of thumb?

49 59 59 60 60 60 61 61 61 61 61 61 61 61 61 61

62 62 62 62 62 62 62 62 63 63 63 63 63 64 64 64

64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64

64 65 65 65 65 65 65 65 65 65 65 65 65 65 65 65

65 65 65 65 65 66 66 66 66 66 66 66 66 66 66 66

66 67 67 67 67 67 67 67 67 67 68 68 68 68 68 68

69 69 69 69

How many outliers are there in weights? What percentage of the data set is this?

Question 5

Answer 1

Solution to Question 5:

According to the range rule of thumb, the maximum value is given by µ+2*ơ whereas minimum value i given by µ-2*ơ

where µ and ơ are the mean and standard deviation respectively.

So, here Minimum approximate value = 64.3 - 2*2.8 = 58.7

and Maximum approximate value = 64.3 +2*2.8 = 69.9

From the dataset we can see the actual minimum is =49 .

So the difference between the approximate minimum and original minimum is 58.7-49=9.7. Hence approximate minimum is 9.7 units greater than the actual minimum.

In case of maximum, the actual maximum =69

So the difference between the approximate maximum and original maximum is 69.9-69=0.9. Hence approximate maximum is 0.9 units greater than the actual maximum.

According to the range rule of thumb, the range is about four times the standard deviation.

Here Range =Actual Maximum value - Actual Min value= 69-49 =20

Since, Range =4* SD (Range rule of thumb)

Standard deviation approx value by range rule of thumb =Range/4=20/4=5

The actual standard deviation =2.8. So the one estimated by range rule of thumb is greater than actual standard deviaton by (5-2.8)=2.2 units.

The outlier is any value that will fall outside the 2 standard deviations range. So, any point of the data that will fall outside µ-2*ơ and  µ+2*ơ will be considered as an outlier. Here we have µ-2*ơ =58.7 and µ+2*ơ =69.9. We observe from the dataset, the datapoint 49 falls outside the range (58.7, 69.9). So 49 is the only outlier here. There is one outlier in the data of weights.

The percentage of data-set of outliers is: (No of outliers)/(Total no of observations)*100= 1/100*100=1%.