Question

SHOW WORK!!!!! Full credit will not be given for answers only. NOTE: for any question asking you to determine a probability----you MUST write out a probability statement using proper notation!!!! Probabilities should be DECIMAL form and rounded to 4 decimal places.

A company manufactures windows that are inserted into automobiles. Each window has five studs for attaching it. A pullout test is used to determine the force required to pull a stud out of a window (i.e. Destructive testing). Let F be the force required for pulling studs out of position. Thirty observations of F were as follows:

159 156 165 142 160 167
151 160 158 148 165 137
161 147 167 158 151 155
120 140 149 160 157 150
137 138 155 144 147 164

NOTE: you can copy and paste the data set into Excel to make it easier to work with. It is acceptable to use excel and summarize the equations/steps used on your homework paper.

a) Create a stem-and-leaf plot.
b) Calculate the sample mean and sample variance.
c) Determine the 5-number summary.
d) Do any outliers exist in our data? If so, identify them and mathematically justify your answer.
e) Construct a well-labeled modified boxplot.
f) Find the 78th percentile.
g) Does the data set appear to be skewed or symmetric? If it is skewed, decide in what direction. Explain your answer.

Answer 1

a.)

Stem and Leaf plot is a plot used for data representation in which numbers are split into the stem part and leaf part.

So, we get:

STEM	LEAF
12	0
13	7,7,8
14	0,2,4,7,7,8,9
15	0,1,1,5,5,6,7,8,8,9
16	0,0,0,1,4,5,5,7,7

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b.)

Sample mean =

Now,

Sample variance =

120	-32.27	1041.13777777778
137	-15.27	233.07111111111
137	-15.27	233.07111111111
138	-14.27	203.53777777778
140	-12.27	150.47111111111
142	-10.27	105.40444444445
144	-8.27	68.33777777778
147	-5.27	27.73777777778
147	-5.27	27.73777777778
148	-4.27	18.20444444444
149	-3.27	10.67111111111
150	-2.27	5.13777777778
151	-1.27	1.60444444444
151	-1.27	1.60444444444
155	2.73	7.47111111111
155	2.73	7.47111111111
156	3.73	13.93777777778
157	4.73	22.40444444444
158	5.73	32.87111111111
158	5.73	32.87111111111
159	6.73	45.33777777778
160	7.73	59.80444444444
160	7.73	59.80444444444
160	7.73	59.80444444444
161	8.73	76.27111111111
164	11.73	137.67111111111
165	12.73	162.13777777778
165	12.73	162.13777777778
167	14.73	217.07111111111
167	14.73	217.07111111111

So,

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c.)

The five-number summary is Maximum, First quartile, Median, 3rd quartile, Maximum.

On looking at the stem and leaf plot we can see that the Minimum value is 120 and the maximum value is 167

Now, we have a total of 30 observations, so we get two middlemost observations at the place of 15th and 16th. So, we take an average of both numbers which will give the median.

STEM	LEAF
12	0
13	7,7,8
14	0,2,4,7,7,8,9
15	0,1,1,5,5,6,7,8,8,9
16	0,0,0,1,4,5,5,7,7

Thus, numbers are 155 and 155, thus the average is which is the median.

Now, the First quartile is the middle value of the lower half of the data. The number of observations in the lower half of the data is 14. So, again we get two middlemost observations at the position of 7th and 8th and then take the average.

So, the first quartile is

Similarly, the 3rd quartile is the middle value of the upper half of the data.

So, we get:

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d.)

For checking the outlier we use:

If an observation is greater than is the outlier

If an observation is less than is the outlier.

Here, IQR is the interquartile range which is calculated as Q3 - Q1 = 160-145.5 = 14.5

So,

and

Thus, the outlier is 120.

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