prob

AIM:
Write the R programming code for computing the mean, median, mode, quartile deviation, variance,
standard deviation, co-efficient of variation, first four moments about the mean and Pearson’s coefficients for the following frequency distribution
KEY WORDS:
Example:
M = Mean
MD = Median
Mode = mode
Cl = cumulative frequency
H = height of the class
F = frequency
V = variance
SD = standard deviation
CV = coefficient of variation
B1 & B2 = Pearson’s Coefficients
M1,M2,M3,M4 = Moments about mean
R-CODE:
h=10
x=seq(175,245,h)
f=c(53,68,85,92,100,95,70,28)
N=sum(f)
 Reg. No: XXXXX
 Name: YYYYY
M=(sum(x*f))/sum(f)
M
cl=cumsum(f)
cl
ml=min(which(cl>=N/2))
ml
F=f[ml]
F
c=cl[ml-1]
c
l=midx[ml]-h/2
MD=l+(((N/2)-c)/f)*h
MD
m=which(f==max(f))
m
fm=f[m]
fm
f1=f[m-1]
f2=f[m+1]
f1
f2
l=midx[m]-h/2
l
mode=l+((fm-f1)/(2*fm-f1-f2))*h
mode
var=((1/N)*(sum(x^2*f)))-((1/N)*(sum(x*f)))^2
var
SD=sqrt(var)
SD
CV=100*(SD/M)
CV
 Reg. No: XXXXX
 Name: YYYYY
M1=sum(((x-M)^1)*f)/N
M2=sum(((x-M)^2)*f)/N
M3=sum(((x-M)^3)*f)/N
M4=sum(((x-M)^4)*f)/N
B1=(M3^2)/(M2^3)
B2=(M4)/(M2^2)
B1
B2
G1=sqrt(B1)
G2=B2-3
G1
G2
________________________________________________________________________________________________
DIGITAL ASSIGNMENT – 3
PROBABILITY AND STATISTICS LAB
Course Code: BMAT202P
NAME :-Muhammad Idris
REG NO – 21BCE37776
CLASS NUMBER: VL2022230503586
LAB SLOT: L5 + L6
AIM :- SOLVING THE GIVEN QUESTIONS USING Binomial, poisson and Normal distributions.
21BCE3695
Code
It is known that probability of an item produced by a certain machine will be defective is
0.05. If the produced items are sent to the market in packets of 20, then write down the
R code to find the number of packets containing at least, exactly and at most 2 defective
items in a consignment of 1000 packets.
# Exp 3 - Binomial, poisson and Normal distributions
# Binomial Distribution
n = 20
p = 0.05
q = 1 - p
N = 1000
#1. Number of items containing at least 2 defective items in a consignment
k = 2
N1 = round(N*(1 - pbinom(k-1,n,p)))
N1
#2. Number of packets containing exactly 2 defective itens in a
consignment k = 2
N2 = round(N*dbinom(k,n,p))
N2
#3. Number of packets containing at most 2 defective items in a consignment
of k = 2
N3 = round(N*(pbinom(k,n,p)))
N3
21BCE3695
Output:-
Conclusion
Hence,
The probability of packets containing at least 2 defective items is – 264
The probability of packets containing exactly 2 defective items is – 189
The probability of packets containing at most 2 defective items is – 925
21BCE3695
A car hire firm has 2 cars which it hires out day by day. The number of demands for a
car on each day follows a Poisson distribution with mean 1.5. Write down the R code to
compute the proportion of days on which (i). neither car is used, (ii). at most one car is
used and (iii). some demand of car is not fulfilled
Code:-
# Poissons Distribution
lam = 1.5
#1. Propotional of days on which neither car is used
x = 0
P1 = dpois(x,lam)
P1
#2. Propotional of days on which atmost one car is
used x = 1
P2 = ppois(x,lam)
P2
#3. Propotional of days on which some demands of car is not fullfilled
x = 2
P3 = 1 - ppois(x,lam)
P3
21BCE3695
Output:-
Conclusion:-
Hence from the analysis,
The proportion of days on which neither car is used – 0.2231302 The
proportion of days on which at most 1 car is used – 0.5578254
The proportion of days on which some demand of car is not fullfilled – 0.1911532.
21BCE3695
The local corporation authorities in a certain city install 10,000 electric lamps in the
streets of the city with the assumption that the life of lamps is normally distributed. If
these lamps have an average life of 1,000 burning hours with a standard deviation of 200
hours, then write down the R code to calculate the number of lamps might be expected
to fail in the first 800 burning hours and also the number of lamps might be expected to
fail between 800 and 1,200 burning hours.
Code:-
# Normal distribution
Mu = 1000
SD = 200
N = 10000
#1. Number of lamps might be expected to fail in the first 800 burning
hours x1 = 800
N1 = round(N*pnorm(x1,Mu,SD))
N1
#2. Number of lamps might be expected to fail between 800 to 1200 burning hours
x1 = 800
x2 = 1200
N2 = round(N*(pnorm(x2,Mu,SD)-pnorm(x1,Mu,SD)))
N2
21BCE3695
Output:-
Conclusion:-
Hence,
The number of lamps might be expected to fail in the first 800 burning hours – 1587
The number of lamps might be expected to fail in between 800 to 1200 burning hours – 6827


NAME : Muhammad Idris
REG.NO : 21BCE3776
SUBJECT CODE : BMAT202P
SUBJECT TITLE : Probability and Statistics LAB
SLOT : L5+L6
SEMESTER : Winter Semester 2022-2023 GUIDED
BY : Dr. MANIMARAN A
Question 1)
Experience has shown that 20% of a manufactured product is of top quality. In one day’s
production of 400 articles, only 50 are of top quality. Write down the R programming
code to test whether the production of the day chosen is a representative sample at 95%
confidence level.
p_hat <- 50/400
p_null <- 0.2
n <- 400
z_score <- (p_hat - p_null) / sqrt(p_null * (1 - p_null) / n)
p_value <- 2 * pnorm(-abs(z_score))
alpha <- 0.05
if (p_value < alpha) {
 cat("Reject the null hypothesis, the production of the day chosen is not a representative sample.")
} else {
 cat("Fail to reject the null hypothesis, the production of the day chosen is a representative sample.")
}
Output:
Reject the null hypothesis, the production of the day chosen is not a representative sample.
Question 2)
Before an increase in excise duty on tea, 800 people out of a sample of 1000 were
consumers of tea. After the increase in duty, 800 people were consumers of tea in a
sample of 1200 persons. Write down the R programming code to test whether the
significant decrease in the consumption of tea after the increase in duty at 1 % level of
significance.
Print (“H0:= p1 = p2”)
Print(“H1:= p1 != p2”)
Alpha = 0.01
ZTab = gnorm(1-alpha)
ZTab
N1 = 1000
N2 = 1200
P1 = 800/1000
P2 = 800/1200
P = ((n1 * p1) + (n2 * p2))/(n1 + n2)
Q = 1 – P
ZCal = ((p1 – p2) / sqrt(P * Q * (1/n1) + (1/n2)))
ZCal
If (abs(ZCal) < abs(ZTab)) {
Print (“H0 is accepted and H1 is rejected”)
} else {
Print (“H0 is accepted and H1 is rejected”)
}
Question 3)
A sample of 900 items is found to have a mean of 3.47 cm. Write down the R
programming code to test whether it can be reasonably regarded as a simple sample
from a population with mean 3.23 cm and SD 2.31 cm at 99% level of confidence.
Print (“H0:= x0 = Mu”)
Print (“H1:= x0 !- Mu”)
Alpha = 0.01
ZTab = gnorm(1 – alpha)
ZTab
Mu = 3.23
Sigma = 2.31
N = 900
X0 = 3.47
ZCal = (x0 – Mu)/ ( Sigma/ sqrt(n))
ZCal
If (abs(ZCal) < abs(ZTab)) {
Print (“H0 is accepted and H1 is rejected”)
} else {
Print (“H0 is rejected and H1 is accepted”)
}
Question 4)
The average mark scored by 32 boys is 72 with a standard deviation of 8, while that for
36 girls is 70 with a standard deviation of 6. Write down the R programming code to
test whether the boys are performing better than girls on the basis of average mark at
5 % level of significance.

Print (“H0 := x1 = x2”)
Print (“H1 := x1 != x2”)
Alpha = 0.05
ZTab = gnorm(1 – alpha)
ZTab
N1 = 32
N2 = 36
X1 = 72
X2 = 70
s1 = 8
S2 = 6
ZCal = (x1-x2) / sqrt(((s1^2) / n1) + ((s2 ^ 2)/n2))
ZCal
If(abs(ZCal) < abs(ZTab)) {
Print (“H0 is accepted and H1 is rejected”)
} else {
Print (“H0 is rejected and H1 is accepted”)
}




NAME : Muhammad Idris
REG.NO : 21BCE3776
SUBJECT CODE : MAT202P
SUBJECT TITLE : PROBABILITY AND STATISTICS LAB
LAB SLOT : L5+L6
SEMESTER : Winter Semester 2022-2023
GUIDED BY : Dr. MANIMARAN A
AIM:
Finding the test statistic in t test, paired test and F test
KEY WORDS:
Mean
Var=Variance
QN 1)
A random sample of 10 boys with the following IQs: 70, 120, 110, 101, 88, 83,
95, 98, 107, and 100. Write down the R programming code to test whether
the data support the assumption of a population mean IQ of 100 at 5 % level
of significance
R-CODE:
print("H0 := X0=Mu")
print("H1 := x0!=Mu")
alpha=0.05
Mu=100
n=10
x=c(70,120,110,101,88,83,95,98,107,100)
x0=mean(x)
s=sqrt(var(x))
tTab=qt((1-alpha),(n-1))
tTab
tCal=(x0-Mu)/(s/sqrt(n-1))
tCal
if(abs(tCal)<abs(tTab)){
print("H0 is accepted and H1 is rejected")
} else {
print("H0 is rejected and H1 is accepted")
}
t.test(x,mu=Mu)
CONCLUSION:
tTab=1.833113
tCal=-0.5885024
H0 is accepted and H1 is rejected
OUPUT SIMULATION AND R-CODE:
QN2)
The mean height and the standard deviation height of 8 randomly chosen soldiers are
166.9 cm and 8.29 cm respectively. The corresponding values of 6 randomly chosen sailors
are 170.3 cm and 8.50 cm respectively. Write down the R programming code to test
whether the soldiers are shorter than the sailors on the basis of average height.
R-CODE:
print("H0 := x1=x2")
print("H1 := x1!=x2")
alpha=0.01
n1=8
n2=6
x1=166.9
x2=170.3
s1=8.29
s2=8.50
Sigma=sqrt(((n1*s1^2)+(n2*s2^2))/(n1+n2-2))
tTab=qt((1-alpha),(n1+n2-2))
tTab
tCal=(x1-x2)/(Sigma*sqrt((1/n1)+(1/n2)))
tCal
if(abs(tCal)<abs(tTab)){
print("H0 is accepted and H1 is rejected")
} else {
print("H0 is rejected and H1 is accepted")
}
CONCLUSION:
tTab=2.680998
tCal=-0.6954801
H0 is accepted and H1 is rejected
RCODE SIMULATION AND OUTPUT:
Q3)
The following data relate to the marks obtained by 11 students in two sets, one held at the
beginning of a year and the other at the end of the year after intensive coaching. Write down the
R programming code to test whether the data indicate that the students have benefited by
coaching at 5 % level of significance?
Test I : 19 23 16 24 17 18 20 18 21 19 20
Test II : 17 24 20 24 20 22 20 20 18 22 19
RCODE:
print("H0 := x1=x2")
print("H1 := x1!=x2")
alpha=0.05
S1=c(19,23,16,24,17,18,20,18,21,19,20)
S2=c(17,24,20,24,20,22,20,20,18,22,19)
D=S1-S2
d=mean(D)
n=length(D)
s=sqrt(var(D))
tTab=qt((1-alpha),(n-1))
tTab
tCal=d/(s/sqrt(n-1))
tCal
if(abs(tCal)<abs(tTab)){
print("H0 is accepted and H1 is rejected")
} else {
print("H0 is rejected and H1 is accepted")
}
CONCLUSION:
tTab=1.812461
tCal=-1.313064
H0 is accepted and H1 is rejected
RCODE AND SIMLUATION:
Q.4) Two random samples drawn from two normal populations with the following
observations.
Sample I : 21 24 25 26 27
Sample II : 22 27 28 30 31 36
Write down the R programming code to test whether the two populations have the same
variance at 5 % level of significance.
RCODE:
print ("HO := s1=s2")
print ("H1 := s1!=s2")
alpha=0.05
s1=c(21, 24, 25, 26, 27)
s2=c(22, 27, 28, 30, 31, 36)
n1=length(s1)
n2=length(s2)
x1=mean(s1)
x2=mean(s2)
s1=sqrt(var(s1))
s2=sqrt(var(s2) )
Var1=(n1/(n1-1))*(s1^2)
Var2=(n2/(n2-1))*(s2^2)
FTab=qf((1-alpha),(n1-1),(n2-1))
FTab
if (Var1>Var2) {
FCal=Var1/Var2
} else {
FCal=Var2/Var1
}
FCal
if(abs(FCal)<abs(FTab)) {
print ("HO Is accepted and H1 is rejected")
} else {
print ("HO Is rejected and H1 is accepted")
}
CONCLUSION:
FTab=5.192168
FCal=3.912453
HO Is accepted and H1 is rejected
OUTPUT AND RCODE SIMULATION:


Sss
NAME : Muhammad Idris
REG.NO : 21BCE3776
SUBJECT CODE : BMAT202P
SUBJECT TITLE : Probability and Statistics Lab
LAB SLOT : L5+L6
SEMESTER : Winter Semester 2022-2023
GUIDED BY : Dr. MANIMARAN A
AIM:
Finding Measures of Central Tendency, Dispersion, Skewness and Kurtosis for the
given data.
KEY WORDS:
Example:
n=Sample Size
alpha= Level of Significance
OF= Given values
Tot= Total of the given values
EV= Calculated Mean
EF= Mean Frequency
ChiTab = Tabular Value
ChiCal = Calculated Value
Dim = Dimensions
. The following table gives the number of fatal road accidents that occurred during the 7
days of a week. Write down the R programming code to test whether the accidents are
uniformly distributed over the week at 95 % level of confidence.
R-CODE:
message(sprintf("80 := The accidents are uniformly distributed over the week"))
message(sprintf("80 := The accidents are not uniformly distributed over the
week"))
alpha=0.05
OF=c(8, 14, 16, 12, 11, 14, 9)
Tot= sum(OF)
n=length(OF)
EV=Tot/n
EF=rep(EV,n)
ChiTab=qchisq((1-alpha),(n-1))
ChiTab
ChiCal=sum(((OF-EF)^2)/EF)
ChiCal
if(abs(ChiCal)<abs(ChiTab)){
message(sprintf("H0 is accepted and H1 is rejected"))
}else{
message(sprintf("H0 is rejected and H1 is accepted"))
}
message(sprintf("H0 := There is no association between gender and attitude"))
message(sprintf("H0 := There is an association between gender and attitude"))
A total number of 3759 individuals were interviewed according to gender and decision in
a public opinion survey on a political proposal with the results as in the following table.
Write down the R programming code to test the hypothesis that there is no association
between gender and attitude 5 % level of significance.
Favoured Opposed Undecided
Male 1154 475 243
Female 1103 442 342
alpha=0.05
OF=matrix(c(1154, 475, 243, 1103, 442, 342),nrow=2,ncol=3)
Dim=dim(OF)
m=Dim[1]
n=Dim[2]
EF=matrix(nrow=m,ncol=n)
for(i in 1:Dim[1])
{
for(j in 1:Dim[2])
{
EF[i,j]=round(sum(OF[i,])*sum(OF[,j])/sum(OF))
}
}
ChiTab=qchisq((1-alpha),(m-1)*(n-1))
message(sprintf("Tabulated Chi-Square Value = %0.4f",ChiTab))
ChiCal=sum(((OF-EF)^2)/EF)
message(sprintf("Calculated Chi-Square Value = %0.4f",ChiCal))
if((abs(ChiCal)<abs(ChiTab))){
message(sprintf("H0 is accepted and H1 is rejected"))
message(sprintf("Attributes are independent"))
} else{
message(sprintf("H0 is rejected and H1 is accepted"))
message(sprintf("Attributes are nit independent"))
}
A random sample is selected from each of 3 makes of ropes (Type 1, Type 2 and Type
3) and their breaking strength (in certain units) are measured with the results in the
following table.
Type 1 : 70 72 75 80 83
Type 2 : 60 65 57 84 87 73
Type 3 : 100 110 108 112 113 120 107
Write down the R programming code to test whether the breaking strengths of the ropes
differ significantly at 5 % level of significance.
#CRD - ANOVA for 1 way classification
D1=c(70, 72, 75, 90, 83)
D2=c(60, 65, 57, 84, 87, 73)
D3=c(100, 110, 108, 112, 113, 120, 107)
Data=c(D1,D2,D3)
Data
Type=c(rep("Type1",length(D1)),rep("Type2",length(D2)),rep("Type3",length(D
3)))
Type
ANOVA1=aov(Data~Type)
summary(ANOVA1)
#RBD - ANOVA for 2 way classification
Data=c(36, 36, 21, 35, 28, 29, 31, 32, 26, 28, 29, 29)
Data
Season=c(rep("Summer",4),rep("Winter",4),rep("Monsoon",4))
Season
Salesman=c(rep(c("SalesA","SalesB","SalesC","SalesD"),3))
Salesman
ANOVA2=aov(Data~(Season+Salesman))
summary(ANOVA2)
#LSD - ANOVA for 3 way classification
Data=c(16, 17, 20, 16, 21, 15, 15, 12, 13)
Data
Day=c(rep("Day1",3),rep("Day2",3),rep("Day3",3))
Day
Engine=c(rep(c("Eng1","Eng2","Eng3"),3))
Engine
Burner=c("B1","B2","B3","B2","B3","B1","B3","B1","B2")
Burner
ANOVA3=aov(Data~(Day+Engine+Burner))
summary(ANOVA3)
CONCLUSION:
1. H0 is accepted H1 is rejected
2. H0 is rejected and H1 is accepted
The attributes are not independent
3.
I. CRD:
II. RBD:
III. LSD: