Question

Please use R programming to answer follow questions:

Given that X has a normal distribution with mean 18 and standard deviation 2.5, find

a. P(X < 15);

b. the value k such that P(X < k) = 0.2236;

c. the value k such that P(X > k) = 0.1814;

d. P(17 < X < 21).

Answer 1

#Given that X has a normal distribution with mean 18 and standard deviation 2.5,

# a. P(X < 15)

# mean m=18 and sd s=2.5
m <- 18
s <- 2.5
xbar <- 15
z <- (xbar-m)/s
#if z is positive then
if(z>=0)
{
res=pnorm(abs(z))
}
#if z is gegative then
if(z<0)
{
res=pnorm(-abs(z))
}
cat("P(X < 15) = " , res,"\n")

#b. the value k such that P(X < k) = 0.2236

#here p value is 0.2236
p<-0.2236
# z value will be generated
z<-qnorm(p)
# value of k calculation
k<-(z*s+m)
cat("The value of k for [P(X < k) = 0.2236 is ] : " , k,"\n")

#c. the value k such that P(X > k) = 0.1814

#as X>k so here p value will be 1-0.1814
p<-(1-0.1814)
# z value will be generated
z<-qnorm(p)
# value of k calculation
k<-(z*s+m)
cat("The value of k for [P(X > k) = 0.1814 is ] : " , k,"\n")

#d. P(17 < X < 21).

xbar <- 17
# z1 value will be generated
z1<- (xbar-m)/s
xbar <- 21
# z2 value will be generated
z2<- (xbar-m)/s
#if z1 is positive then
if(z1>=0)
{
res1<-pnorm(abs(z1))
}
#if z1 is negative then
if(z1<0)
{
res1<-pnorm(-abs(z1))
}
#if z2 is positive then
if(z2>=0)
{
res2<-pnorm(abs(z2))
}
#if z2 is negative then
if(z2<0)
{
res1<-pnorm(-abs(z2))
}
res=res2-res1
cat("P(17<X<21) = " , res,"\n")

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OUTPUT