#The weights of steers in a herd are distributed normally. The variance is 90,000 and the
#mean steer weight is 800 lbs. Find the probability that the weight of a randomly
#selected steer is between 1040 and 1460 lbs.
#mean=800
#sd=sqrt(90000)
#sd
PW<-pnorm(1040, mean=800, sd=sqrt(90000), lower.tail = FALSE) -
pnorm(1460, mean=800, sd=sqrt(90000), lower.tail = TRUE)
#ggplot(mapping = aes(Weigh = PW)) + stat_qq_point(size = 2)
PW
## [1] -0.7742412
qqnorm(PW,
ylab="Weigh",
xlab="Herd",
main="weights of steers in a herd")
qqline(PW)
PW
## [1] -0.7742412
RPW<-round(PW,digits = 4)
RPW
## [1] -0.7742
#- Above code is mine
#I am now trying with your (AS) solution
#x<-c(1040,1460)
sum(dnorm(x=1040:1460,mean=800,sd=300))
## [1] 0.198494
pnorm(q=1460, mean=800, sd=sqrt(90000)) -
pnorm(q=1040, mean=800, sd=sqrt(90000))
## [1] 0.197952
#continous distribution, not discrete.
pnorm(q=1040, mean=800, sd=sqrt(90000), lower.tail = FALSE) -
pnorm(q=1460, mean=800, sd=sqrt(90000), lower.tail = FALSE)
## [1] 0.197952
round(x=(pnorm(q=1460, mean=800, sd=sqrt(90000)) -
pnorm(q=1040, mean=800, sd=sqrt(90000))),digits=4)
## [1] 0.198
?seq()
## starting httpd help server ... done
x <- seq(from = -4,
to = 4,
length = 1000) * 300 + 800
#create a vector of values that shows the height of the probability distribution for each value in x
y <- dnorm(x = x, # argument = vector x created by seq() function
mean = 800,
sd = 300
)
#plot normal distribution with customized x-axis labels
plot(x = x,
y = y,
type = "l",
lwd = 2,
axes = FALSE,
xlab = "",
ylab = ""
)
sd_axis_bounds <- 5
axis_bounds <- seq(from = -sd_axis_bounds *300 + 800,
to = sd_axis_bounds * 300 + 800,
by = 300)
axis(side = 1, at = axis_bounds, pos = 0)
#The diameters of ball bearings are distributed normally. The mean diameter is 106
#millimeters and the standard deviation is 4 millimeters. Find the probability that the
#diameter of a selected bearing is between 103 and 111 millimeters.
#mean=106
#sd=4
#P(X) between 103 and 111
data<-103:111
data
## [1] 103 104 105 106 107 108 109 110 111
dnorm(x=data,mean=106,sd=4)
## [1] 0.07528436 0.08801633 0.09666703 0.09973557 0.09666703 0.08801633 0.07528436
## [8] 0.06049268 0.04566227
sum(dnorm(x=data,mean=106,sd=4))
## [1] 0.725826
PW<-pnorm(q=111, mean=106, sd=4)
- pnorm(q=103, mean=106, sd=4)
## [1] -0.2266274
PW
## [1] 0.8943502
??diff
diff(x=pnorm(c(103,111),mean=106,sd=4))
## [1] 0.6677229
pnorm(q=103, mean=106, sd=4,lower.tail = F)
## [1] 0.7733726
- pnorm(q=111, mean=106, sd=4,lower.tail = F)
## [1] -0.1056498
#The lengths of nails produced in a factory are normally distributed with a mean of 3.34
#centimeters and a standard deviation of 0.07 centimeters. Find the two lengths that
#separate the top 3% and the bottom 3%. These lengths could serve as limits used to
#identify which nails should be rejected.
mean=3.34
sd=0.07
qnorm(p=.03,mean=3.34,sd=0.07)
## [1] 3.208344
qnorm(p=0.97,mean=3.34,sd=0.07)
## [1] 3.471656
round(x=c(qnorm(p=0.03,mean=3.34,sd=0.07),qnorm(p=0.97,mean=3.34,sd=0.07)),digits=2)
## [1] 3.21 3.47
# bottom 3% and the top 3%.
qnorm(p=c(.03,.5,0.97),mean=3.34,sd=.07)
## [1] 3.208344 3.340000 3.471656
x <- seq(from = -4, to = 4, length = 1000) * .07 +3.34
#create a vector of values that shows the height of the probability distribution
#for each value in x
y <- dnorm(x, mean=3.34, sd=.7)
#plot normal distribution with customized x-axis labels
plot(x,y, type = "l", lwd = 2, axes = FALSE, xlab = "", ylab = "")
sd_axis_bounds = 5
axis_bounds <- seq(from = -sd_axis_bounds *.07 + 3.34,
to = sd_axis_bounds *.07 + 3.34,
by = .07)
axis(side = 1, at = axis_bounds, pos = 0)
