MAS 183 2017 final Exam Answers
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School
Murdoch University *
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Course
183
Subject
Mathematics
Date
Apr 3, 2024
Type
docx
Pages
10
Uploaded by SuperDeerPerson880
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Answers to Practise Test
Q 1
Volume counted = 9 mm
2
x 0.1 mm = 0.9 mm
3
= 0.9 uL.
a.
What is the cell count per microlitre (µL)?
= 20 x 17 per 0.9 uL
= 340 cells /0.9 uL
= 378 cells /uL
b.
What is the cell count per millilitre (mL)?
= 378 x 10
3
/ mL
or
3.78 x 10
5
cells / mL
c.
What is the cell count per Litre?
3.78 x 10
8
cells / L
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Answers to Practise Test
Q 2
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Answers to Practise Test
Q 3
correlation coefficient =
0.63
As the correlation coefficient (r=0.63) is significantly larger than 0.3, I conclude that there is a significant positive correlation between the counts in chambers 1&2 of each slide.
Note: Using XLSTAT, there is significant evidence of positive correlation (p
value = 5.4E-5
200
300
400
500
600
700
800
300
350
400
450
500
550
600
650
700
R² = 0.39
Count (Chamber 1)
Count (Chamber 2)
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Answers to Practise Test
Q 4
3
0
1
3
1
4
7
3
2
2
3
7
3
3
0
2
8
9
3
4
1
2
6
9
3
5
1
1
1
2
2
2
3
4
3
6
2
5
6
8
3
7
0
5
7
7
8
3
8
0
5
5
3
9
3
8
4
0
4
7
9
4
1
4
2
4
3
9
Q 5
Median = Q 6
Provide a 5 number summary of the data:
Min
= 301
Q
1
= 340.5
(Using Quartile.inc)
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Answers to Practise Test
Q
2
= 352.5
Q
3
= 377.25
Max
= 439
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Answers to Practise Test
Q 7
250
270
290
310
330
350
370
390
410
430
450
Diameter (µm)
Q 8
Lower Limit = 340.5 – 1.5*36.75 = 285
Upper Limit = 377.25 + 1.5 * 36.75 = 432
(
IQR = 377.25 – 340.5 = 36.75)
Does all the data lie between the lower and upper limits? NO, 439 is an outlier
Discuss briefly, the action you would take if you discovered an outlier in the data set.
Note that these are the inner fences; therefore it is not unusual to find a value lying outside. Check with supervisor or colleague as necessary. Proceed with caution as these “points of interest” may belong inside the population. If the outer fences are calculated (Q
3,1
± 3*IQR)– data points lying outside these are highly suspect and should probably/definitely be removed from the data set.
In this case we are advised it should be removed from the data set. DATA POINT NOTED DOWN, THEN ERASED: (439)
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Answers to Practise Test
Q 9
Data point 439 removed from the data set
Diameter (
μm
)
Midpoint
x
i
Tally
Frequency
301-320
310
III
3
321-340
330
IIII II
7
341-360
350
IIII
IIII
II
12
361-380
370
IIII
IIII
10
381-400
390
IIII
4
401-420
410
III
3
Q 10
On the graph paper provided, draw a fully labelled frequency histogram for the data.
310
330
350
370
390
410
430
0
2
4
6
8
10
12
14
Diameter (µm)
Frequency
Q 11
Average = 357 (3sf)
Standard deviation = 26.8 (3sf)
Q 12
CV(%) = 7.5 %
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Answers to Practise Test
Q 13
x
−
t
s
√
n
<
μ
<
x
+
t
s
√
n
x
=
357.3
s
=
26.80
n
=
39
t
=
2.024
357.3
−
2.024
∗
26.8
√
39
<
μ
<
357.3
+
2.024
26.8
√
39
=348.6 < u < 366.0 µm (4sf)
Box and Whisker Plot
329
367
324
319
379
344
371
363
360
356
300
320
340
360
380
400
ppm (Cu)
Q 14
BLUE (bottom) * Larger IQR , Q3 and Max
* Smaller Min, Q1 and Q2
* has less symmetry
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Answers to Practise Test
Which of the two data sets described by the Box and Whisker Plot (above) is more likely to follow a normal distribution – give at least one reason for your answer.
Q 15
Both data sets can be modelled using a normal distribution as they display strong central tendency. However of the two, the top data set:
* is more symmetrical
* The range is smaller suggesting more precision in the measurements
Q 16
Any one of these ……
Central Tendency
Symmetrical distribution
68 / 95 / 99.7% rule
Median = Mode = Mean
Asymptotic to x axis
Q 17
(b) Use the statistical tables to find P(z
-1.25)
P(z
-1.25) = 1 - P(z < -1.25)
= 1 – 0.0401
= 0.9599
Q 18
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Answers to Practise Test
Q 19
Use the statistical tables to find P(-0.87
z
1.26)
P(-0.87
z
1.26)
=
P(z
1.26) - P( z
-0.87)
=
0.8962 - 0.1922
=
0.7040
Q 20
Find the value of z (below).
z = 1.96
Q 21
Measurement Measurement
Uncertainty
% Uncertainty
Length (mm) (L) 4.51
0.02
0.44%
Width (mm) (W)
5.50
0.02
0.36%
Depth (mm) (D)
12.6
0.02
0.16%
Volume (LDW)
313 (3sf)
±1.9 cm
3
0.60%
Calculate Right column results first.
Calculate Volume = LDW next.
Then calculate the measurement uncertainty on the volume as 0.6% of 313