The data displayed below shows the shaft diameter for various parts coming off an
assembly line.
a) Using the stem-and-leaf plot below, calculate:
Stem-and-Leaf Display: Diameter
Stem-and-leaf of Diameter N = 15
Leaf Unit = 1.0
5 234
5 5689
6 24
6 567
7 12
7
8
8 9
MEAN = 943/15 = 62.87
_______
MEDIAN = 62
_______
RANGE = 52 to 89
_______
STANDARD DEVIATION = 9.33714
b) Describe the distribution of values within the stem-leaf plot.
The stem and leaf display is unimodel and skewed to right. there is 1 outlier of 89.
b) What is the best measure of centre and spread for the shaft diameters. Explain your you choice.
The best measure of centre and spread is the median. The mean is pulled to right by outlier of 89. The median is more realistic as most numbers are closer to it.
c)
Identify any outlier(s) by performing the appropriate calculations. This is part of question I don not know how to calculate.
d) [FONT=Arial,Bold][FONT=Arial,Bold]Suppose that there is an error in one of our data points. The “89” was actually a “79”. [/FONT][FONT=Arial,Bold][FONT=Arial,Bold][/FONT][/FONT]Indicate whetherthis mistake would make the values in part (a) greater, lesser, or almost unchanged once we correct the data:
Mean would be 62.2 and Standard Deviation would be 7.92104 so the the values would be lessor.
I believe my calculations for the above questions are correct, could you please explain how to calculate the outlier of 89.
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assembly line.
a) Using the stem-and-leaf plot below, calculate:
Stem-and-Leaf Display: Diameter
Stem-and-leaf of Diameter N = 15
Leaf Unit = 1.0
5 234
5 5689
6 24
6 567
7 12
7
8
8 9
MEAN = 943/15 = 62.87
_______
MEDIAN = 62
_______
RANGE = 52 to 89
_______
STANDARD DEVIATION = 9.33714
b) Describe the distribution of values within the stem-leaf plot.
The stem and leaf display is unimodel and skewed to right. there is 1 outlier of 89.
b) What is the best measure of centre and spread for the shaft diameters. Explain your you choice.
The best measure of centre and spread is the median. The mean is pulled to right by outlier of 89. The median is more realistic as most numbers are closer to it.
c)
Identify any outlier(s) by performing the appropriate calculations. This is part of question I don not know how to calculate.
d) [FONT=Arial,Bold][FONT=Arial,Bold]Suppose that there is an error in one of our data points. The “89” was actually a “79”. [/FONT][FONT=Arial,Bold][FONT=Arial,Bold][/FONT][/FONT]Indicate whetherthis mistake would make the values in part (a) greater, lesser, or almost unchanged once we correct the data:
Mean would be 62.2 and Standard Deviation would be 7.92104 so the the values would be lessor.
I believe my calculations for the above questions are correct, could you please explain how to calculate the outlier of 89.
[/FONT]