The time to failure in hours of an electronic component subjected to an accelerated life test is shown in Table 3E.1. To accelerate the failure test, the units were tested at an elevated temperature.
Calculate the sample average and standard deviation.
Construct a histogram.
Construct a stem-and-leaf plot.
Find the sample median and the lower and upper quartiles.
data38 <- c(
127,124,121,118,
125,123,136,131,
131,120,140,125,
124,119,137,133,
129,128,125,141,
121,133,124,125,
142,137,128,140,
151,124,129,131,
160,142,130,129,
125,123,122,126
)
length(data38)
## [1] 40
mean(data38)
## [1] 129.975
Berdasarkan hasil perhitungan menggunakan R, diperoleh sample average sebesar 129.975 jam.
sd(data38)
## [1] 8.914084
Sample standard deviation yang diperoleh adalah 8.914084 jam.
hist(data38,
breaks = seq(115,165,by=5),
main = "Histogram Electronic Component Failure Time",
xlab = "Failure Time (hours)",
ylab = "Frequency")
Berdasarkan histogram, sebagian besar waktu kegagalan komponen berada
pada kisaran 120–140 jam. Terdapat beberapa pengamatan dengan waktu
kegagalan yang relatif tinggi.
stem(data38)
##
## The decimal point is 1 digit(s) to the right of the |
##
## 11 | 89
## 12 | 0112334444
## 12 | 555556788999
## 13 | 011133
## 13 | 677
## 14 | 00122
## 14 |
## 15 | 1
## 15 |
## 16 | 0
median(data38)
## [1] 128
quantile(data38)
## 0% 25% 50% 75% 100%
## 118.00 124.00 128.00 133.75 160.00
Berdasarkan hasil perhitungan, diperoleh: - Kuartil bawah (Q1) = 124 jam - Median = 128 jam - Kuartil atas (Q3) = 133.75 jam
# Latihan 3.10
An article in Quality Engineering (Vol. 4, 1992, pp. 487–495) presents viscosity data from a batch chemical process. A sample of these data is presented in Table 3E.3 (read down,then across). (a)Construct a stem-and-leaf display for the viscos ity data. (b)Construct a frequency distribution and histogram. (c)Convert the stem-and-leaf plot in part (a) into an ordered stem-and-leaf plot. Use this graph to assist in locating the median and the upper and lower quartiles of the viscosity data. (d)What are the tenth and ninetieth percentiles of viscosity?
data310 <- c(
13.3,14.9,15.8,16.0,
14.5,13.7,13.7,14.9,
15.3,15.2,15.1,13.6,
15.3,14.5,13.4,15.3,
14.3,15.3,14.1,14.3,
14.8,15.6,14.8,15.6,
15.2,15.8,14.3,16.1,
14.5,13.3,14.3,13.9,
14.6,14.1,16.4,15.2,
14.1,15.4,16.9,14.4,
14.3,15.2,14.2,14.0,
16.1,15.2,16.9,14.4,
13.1,15.9,14.9,13.7,
15.5,16.5,15.2,13.8,
12.6,14.8,14.4,15.6,
14.6,15.1,15.2,14.5,
14.3,17.0,14.6,12.8,
15.4,14.9,16.4,16.1,
15.2,14.8,14.2,16.6,
16.8,14.0,15.7,15.6
)
length(data310)
## [1] 80
stem(data310)
##
## The decimal point is at the |
##
## 12 | 68
## 13 | 1334
## 13 | 677789
## 14 | 0011122333333444
## 14 | 555566688889999
## 15 | 1122222222333344
## 15 | 566667889
## 16 | 011144
## 16 | 56899
## 17 | 0
Berdasarkan stem-and-leaf display, data viscosity tersebar pada nilai sekitar 12,6 hingga 17,0.
breaks310 <- seq(12.5, 17.5, by = 0.5)
freq310 <- hist(data310,
breaks = breaks310,
plot = FALSE)
freq310$counts
## [1] 2 4 8 18 11 17 9 6 5 0
freq_table310 <- data.frame(
Interval = paste(
head(freq310$breaks, -1),
"-",
tail(freq310$breaks, -1)
),
Frequency = freq310$counts
)
freq_table310
## Interval Frequency
## 1 12.5 - 13 2
## 2 13 - 13.5 4
## 3 13.5 - 14 8
## 4 14 - 14.5 18
## 5 14.5 - 15 11
## 6 15 - 15.5 17
## 7 15.5 - 16 9
## 8 16 - 16.5 6
## 9 16.5 - 17 5
## 10 17 - 17.5 0
hist(data310,
breaks = breaks310,
main = "Histogram Viscosity",
xlab = "Viscosity",
ylab = "Frequency")
Berdasarkan histogram, sebagian besar data viscosity berada pada kisaran
14,0 hingga 15,5. Distribusi data terlihat terkonsentrasi di sekitar
nilai tersebut, dengan beberapa pengamatan pada nilai yang lebih rendah
maupun lebih tinggi.
stem(sort(data310))
##
## The decimal point is at the |
##
## 12 | 68
## 13 | 1334
## 13 | 677789
## 14 | 0011122333333444
## 14 | 555566688889999
## 15 | 1122222222333344
## 15 | 566667889
## 16 | 011144
## 16 | 56899
## 17 | 0
Data yang telah diurutkan digunakan untuk membantu menentukan median serta kuartil bawah dan kuartil atas.
median(data310)
## [1] 14.9
Median dari data viscosity adalah 14,9.
quantile(data310)
## 0% 25% 50% 75% 100%
## 12.600 14.300 14.900 15.525 17.000
Berdasarkan hasil perhitungan menggunakan R, diperoleh kuartil bawah (Q1) sebesar 14,3, median sebesar 14,9, dan kuartil atas (Q3) sebesar 15,525.
quantile(data310, probs = c(0.10, 0.90))
## 10% 90%
## 13.70 16.13
Berdasarkan hasil perhitungan menggunakan R, diperoleh persentil ke-10 (P10) sebesar 13,70 dan persentil ke-90 (P90) sebesar 16,13.