Solution of question of midterm exam (STA 201)

Solution 1(b)

q1_b<-c(20.71,20.46,19.04,19.97,21.04,21.15,19.24,20.25,18.73,20.38,22.01,21.59,20.40,21.87)
sum(q1_b) #Sum of X
## [1] 286.84
mean(q1_b) # Mean
## [1] 20.48857
median(q1_b) # Median
## [1] 20.43
table(q1_b) # Mode does not exist
## q1_b
## 18.73 19.04 19.24 19.97 20.25 20.38  20.4 20.46 20.71 21.04 21.15 21.59 
##     1     1     1     1     1     1     1     1     1     1     1     1 
## 21.87 22.01 
##     1     1
quantile(q1_b,0.90) # P90 (90th Percentile)
##    90% 
## 21.786
IQR(q1_b)   # IQR
## [1] 1.0825
sum(q1_b^2)#  Sum of X-square
## [1] 5890.265
sd(q1_b)  # Standard deviation of X
## [1] 1.012346

Solution 2(a)

q2_a<-c(680,     669,   719,    699,    670,    710,    722,    663,    658,    634,
720,    690,    677,    669,    700,    718,    690,    681,    702,    696,
692,    690,    694,    660,    649,    675,    701,    721,    683,    735
)
fd<-cut(q2_a,seq(620,740,20),right = FALSE)
FD<-as.data.frame(table(fd))
colnames(FD)<-c("Class interval", "Frequency")
FD
##   Class interval Frequency
## 1      [620,640)         1
## 2      [640,660)         2
## 3      [660,680)         7
## 4      [680,700)        10
## 5      [700,720)         6
## 6      [720,740)         4
barplot(table(fd),space = 0.001)

hist(q2_a,col = 2,main = "frequency histogram of measurement")

Solution of 2(b)

Hours_TV<-c(    7,  10, 11, 12, 14, 15, 15, 16)
Marks<-c(   85, 100,    90, 95, 84, 75, 75, 90)
plot(Hours_TV,Marks)
abline(lm(Marks~Hours_TV))

Solution 3(b)

library(tidyr)
library(dplyr)
df<-data.frame(Low=c(242,   249,    235,    250,    254,    244,    258,    311,    237,    261), High=c(302,   421,    419,    399,    317,    311,    350,    363,    392,    367))
df
##    Low High
## 1  242  302
## 2  249  421
## 3  235  419
## 4  250  399
## 5  254  317
## 6  244  311
## 7  258  350
## 8  311  363
## 9  237  392
## 10 261  367
df_tidy<-gather(df,key="Vibration_level",value = "Force")
df_tidy
##    Vibration_level Force
## 1              Low   242
## 2              Low   249
## 3              Low   235
## 4              Low   250
## 5              Low   254
## 6              Low   244
## 7              Low   258
## 8              Low   311
## 9              Low   237
## 10             Low   261
## 11            High   302
## 12            High   421
## 13            High   419
## 14            High   399
## 15            High   317
## 16            High   311
## 17            High   350
## 18            High   363
## 19            High   392
## 20            High   367
df_tidy%>%group_by(Vibration_level)%>%
  summarise(Mean=mean(Force),SD=sd(Force))
## # A tibble: 2 x 3
##   Vibration_level  Mean    SD
##   <chr>           <dbl> <dbl>
## 1 High             364.  43.9
## 2 Low              254.  21.7
boxplot(Force~Vibration_level,data =df_tidy,col=rainbow(2) )

Solution 4(a)

Prob(The circuit operates)=0.929258

Prob_row1<-0.9*.8*.7
Prob_row1
## [1] 0.504
Prob_row2<-0.95^3
Prob_row2
## [1] 0.857375
Prob_whole<-1-(1-Prob_row1)*(1-Prob_row2)
Prob_whole
## [1] 0.929258