Calculating the probability of a sample mean using R

Question

I'm in an intro to stats class right now, and have absolutely no idea what's going on. How would I solve the following problem using R?

Let x be a continuous random variable that has a normal distribution with a mean of 71 and a standard deviation of 15. Assuming n/N is less than or equal to 0.05, find the probability that the sample mean, x-bar, for a random sample of 24 taken from this population will be between 68.1 and 78.3,

I'm really struggling on this one and I still have to get through other problems in the same format. Any help would be greatly appreciated!

Answer 1

For R coding this might set you up:

[# Children's IQ scores are normally distributed with a
# mean of 100 and a standard deviation of 15. What
# proportion of children are expected to have an IQ between
# 80 and 120?

mean=100; sd=15
lb=80; ub=120

x <- seq(-4,4,length=100)*sd + mean
hx <- dnorm(x,mean,sd)

plot(x, hx, type="n", xlab="IQ Values", ylab="",
  main="Normal Distribution", axes=FALSE)

i <- x >= lb & x <= ub
lines(x, hx)
polygon(c(lb,x\[i\],ub), c(0,hx\[i\],0), col="red") 

area <- pnorm(ub, mean, sd) - pnorm(lb, mean, sd)
result <- paste("P(",lb,"< IQ <",ub,") =",
   signif(area, digits=3))
mtext(result,3)
axis(1, at=seq(40, 160, 20), pos=0)]

There is also some nice introductory course to R and data analysis by datacamp, this might also come in handy: https://www.datacamp.com/courses/exploratory-data-analysis

And another tutorial on R and statistics: http://www.cyclismo.org/tutorial/R/confidence.html

Answer 2

In terms of the code:

pop_sample <- rnorm(24, 71, 15)
se_pop <- sd(pop_sample)/sqrt(24)
pnorm(78.3, 71, se_pop) - pnorm(68.1, 71, se_pop) # 80%

In term of stats... you should probably refer to stats.stackexchange.com or your professor.