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I'm having trouble solving a normal distribution problem in R. I'm unfamiliar with the syntax and would like some help.
If X~N(2,9), compute
a. P(X>=2)
b. P(1<=X<7)
c. P(-2.5<=X<-1)
d. P(-3<=X-2<3)
You are looking for the pnorm function. This is the normal CDF. So you want to do something like:
# A
1 - pnorm(2, mean = 2, sd = 9) # = 0.5
# B
pnorm(7, mean = 2, sd = 9) - pnorm(1, mean = 2, sd = 9) # = 0.255
I think you can figure out the last two yourself.
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For example, I try to solve for x: 12*exp(-2x/4)=0.5, or x^2+3x^4+9x^5=10, is there any method in R to solve such equations?
Try with uniroot function
r1 <- uniroot(function(x) 12*exp(-x/2) - 0.5, c(-1000, 1000))
r2 <- uniroot(function(x) x^2 + 3*x^4 + 9*x^5 - 10, c(-1000, 1000))
r1$root
[1] 6.356108
r2$root
[1] 0.943561
You have to define:
a function of the type f(x) = 0 (and remove the second member of equality)
an interval within which to search for the solution
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What is a more elegant way to write the below function. I am trying to practice my function development skills, and I am simply trying to manually recreate the CIs of a linear model. I am very well aware of the confint(model) function, but still...
jad<-function(x, y) {
model<-lm(y~x)
std.err<-coef(summary(model))[, 2]
coef.model1<-coef(summary(model))[, 1]
upper.ci<-coef.model1+1.96*std.err
lower.ci<-coef.model1-1.96*std.err
print(upper.ci)
print(lower.ci)
}
What about the code below?
jad <- function(x, y) {
`colnames<-`(
coef(summary(lm(y ~ x)))[,1:2] %*% matrix(c(1, 1.96, 1, -1.96), nrow = 2),
c("upper.ci", "lower.ci")
)
}
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solution of the bit error probability problem
prob=function(E,m) #--- prob is the estimated error probabity for given values of signal to
#---noise ratio E and sample size m
{
stopifnot(E>=0 & m>0) #--- this says that the function won't accept negative values of E and
#---m shoulde be at least 1
n=rnorm(m) #--- this says that n is a random sample of size m from N(0,1)
#---distribution
m=mean(n< -sqrt(E)) #--- this says that m is the proportion of values in n which are less
#---than the negative root of E
return(m) #--- this gives us the value of m, which is the estimated error
#---probability
}
E=seq(0,2,by=0.001)
sam=1000
y=sapply(E,prob,m=sam)
p=10*log10(E)
plot(p, log(y),
main="Graph For The Error Probabilities",
xlab=expression(10*log[10](E)),
ylab="log(Error Probability)",
type="l")
You can try the MATLAB code like below
clc;
clear;
close all;
function y = prob(E,m)
assert(E>=0 & m>0);
n = randn(1,m);
y = mean(n+sqrt(E)<0);
end
E = 0:0.001:2;
sam = 1000;
y = arrayfun(#(x) prob(x,sam),E);
p = 10*log10(E);
plot(p,log(y));
title("Graph For The Error Probabilities");
xlabel('10\log_{10}(E)');
ylabel("log(Error Probability)");
OUTPUT (MATLAB)
OUTPUT (R)
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I am trying to solve a Mathematical equation in one of my geometric modelling problem.
Let's say if I have 2 vectors, A and B, and I have the following equation:
A x B = c (c is a scalar value).
If I know the coordinate of my vector B (7/2, 15/2); and I know the value of c, which is -4.
How can I calculate my vector A, to satisfy that equation (A X B = c) ?
The problem is underdetermined; there isn't a unique such A. I assume that by "multiplication" you mean the cross product.
A = (x,y)
B = (7/2, 15/2)
A×B = x(15/2) - y(7/2)
-4 = (15x-7y)/2
15x - 7y = -8
This gives a line along which points A=(x,y) can lie. Specifically, for any real number t,
x = -1 + 7t
y = -1 + 15t
gives a solution.
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the question is:
x dy/dx = 2y ; y(0)=0
because when i solve this problem the integration constant 'c' gets zero... and i have to find its value in order to calculate a solution to given IVP
Unless I'm mistaken, this question gives c = 0 for y(0) = 0
x*dy/dx = 2y
x*dy = 2y*dx
dy / 2y = dx / x
ln(2y) = ln(x) + c
e^(ln(2y)) = e^(ln(x) + c) = e^ln(x)*e(c)
2y = x + c
solving for y(0) = 0 gives c = 0, as you stated.
Why do you think c must not be 0?