Ship Maintenance ME4704
Ir. Dwi Priyanta, MSE – NIP. 196807031994021003
Assignment 1
By I Made Yoga Sathya Wibawa – NRP. 5019201023
1. The mean tensile strength of a synthetic fiber is an important quality characteristic that
is of interest to the manufacturer. A random sample of 16 specimens is selected, and
their tensile strengths are determined. Determine mean, median, mode, range, standard
deviation as well as variance of the random sample.
a. Mean (𝑥̅ )
𝑛
1
𝑥̅ = ∑ 𝑥𝑖
𝑛
1
𝑥̅ =
1
𝑥(48.89 + 52.07 + 49.29 + 51.66 + 52.16 + 49.72 + 48.00 + 49.96 + 49.20
16
+ 48.10 + 47.90 + 46.94 + 51.76 + 50.75 + 49.86 + 51.57)𝑝𝑠𝑖
𝑥̅ = 49,86 𝑝𝑠𝑖
b. Median
The data of these speciment must be arranged for increasing or decreasing.
Specimen
Strength
(psi)
12
11
7
10
1
9
3
6
15
8
14
16
4
13
2
5
46,94
47,90
48,00
48,10
48,89
49,20
49,29
49,72
49,86
49,96
50,75
51,57
51,66
51,76
52,07
52,16
The median states in the middle of this arranged data. The median of these data is …
(49,72 psi + 49,86 psi)/2 = 49,79 psi
c. Mode
Mode is defined as the value that occurs most often in a set of samples. These value of datas
show once only, there is no value that shows more than once which mean these data has no
mode.
d. Range
Range of these data is defined between the smallest value and the biggest value which
is between 46,94 and 52,16
e. Sample deviation
Sample deviation is defined as positive square root variance of the data.
𝑛
1
𝑆=√
∑(𝑥𝑖 − 𝑥̅ )2
𝑛−1
1
𝑆 = 1,661248
Ship Maintenance ME4704
Ir. Dwi Priyanta, MSE – NIP. 196807031994021003
2. The stem and leaf is one of pictorial statistics that useful for identifying whether there
are any outliers data, and the general shape of the distribution. Plot the data presented
in example 1 using stem-and-leaf.
N
Leaf unit
: 16
: 0,01
Stem
46
47
48
49
50
51
52
Leaf
94
9
0 1 89
2 29 72 86 96
75
57 66 76
07 16
Table 2.1 Stem and leaf
3. A box plot displays three quartiles, the minimum, and the maximum of the data on a
rectangular box, aligned horizontally and vertically. The box encloses the interquartile
range with the left (or lower) line at the first quartile Q1 and the right (or upper) line at
the third quartile Q3. A line is drawn through the box at the second quartile Q2. A line
at either and either end extends to the extreme values. Plot the data presented in example
1 using the box plot.
Arangged specimen
12
11
7
10
1
9
3
6
Q1
15
8
14
Q2
Q3
𝑥10 + 𝑥1
= 48,49 𝑝𝑠𝑖
2
𝑥6 + 𝑥15
𝑄2 =
= 49,79 𝑝𝑠𝑖
2
𝑥16 + 𝑥4
𝑄3 =
= 51,61 𝑝𝑠𝑖
2
𝑄1 =
Parameter
First Quartile
Second Quartile
Third Quartile
Minimum Value
Maximum Value
16
Value
48,49 psi
49,79 psi
51,61 psi
46,94 psi
52,16 psi
Tabel 3.1 couples of properties
4
13
2
5
Ship Maintenance ME4704
Ir. Dwi Priyanta, MSE – NIP. 196807031994021003
4. The same compressor is manufactured at three plants. Plant 1 makes 70% of the
requirement and plant 2 makes 20%. From plant 1, 90% of compressors meet particular
standard, from plant 2 only 80%, whereas from plant 3 the percentage of unit produced
up to standard is 95%. Determine : How many compressors will be up to standard out
of every 100 purchased by customers? Given that the purchased compressor is up to
standard, what is the probability that it was made in plant 2.
As the condition stated above, it can be written like this.
a. On plant 1, 70 works done of 100 requirements. From 70 only 63 works meet
the standard.
b. On Plant 2, 20 works done of 100 requirements. From 20 only 16 works meet
the standard.
c. On Plant 3, 10 works done of 100 requirements. From 10 works, 9 works meet
the standard.
The number of compressors that meet the standard out of every 100 purchased
by customers are 88 compressors.
The probability of the next custumer after knowing this quality is the plant 2
will be replaced with plant 3 due to the efficiency of plant 3 is better than plant
2.
5. A system consists of 17 different components, of which one, component K, has been
identified as the keystone component. The probability that the keystone component is
good is 0.95. If the keystone component is good, then the system succeeds with a
probability 0.9. If not, then the probability of system success drop to 0.8. What is the
probability of system success
𝑆𝑘 = 95
𝐹𝑘 = 5
𝑆𝑠 𝑖𝑓 𝑆𝑘 𝑔𝑜𝑜𝑑 = 90
𝐹𝑠 𝑖𝑓 𝑆𝑘 𝑔𝑜𝑜𝑑 = 10
𝑆𝑠 𝑖𝑓 𝑆𝑘 𝑏𝑎𝑑 = 80
𝐹𝑠 𝑖𝑓 𝑆𝑘 𝑏𝑎𝑑 = 20
90 + 80
𝑆
2
𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = 𝑝 =
=
= 0,85
𝑆 + 𝐹 (90 + 80 + 10 + 20)
2
2
10 + 20
𝑆
2
𝑃(𝑓𝑎𝑖𝑙𝑢𝑟𝑒) = 𝑞 =
=
= 0,15
𝑆 + 𝐹 (90 + 80 + 10 + 20)
2
2
𝑝+𝑞 =1
0,85 + 0,15 = 1