Calculate Proportion in R using Normal Distribution
I was working on statistics using R. Before i do this using R program, i have done it manually. So here is the problem.
A sample of 300 TV viewers were asked to rate the overall quality of television shows from 0 (terrible) to 100 (the best). A histogram was constructed from the results, and it was noted that it was mound-shaped and symmetric, with a sample mean of 65 and a sample standard deviation of 8. Approximately what proportion of ratings would be above 81?
I have answered it manually with this : Pr(X>81)=Pr(Z>(81-65)/8)=Pr(Z>2)=0.0227 So the proportion is 0.023 or 2.3%
I have trouble with how can i do this in R ? I have tried using pnorm(p=..,mean=..,sd=..) but didnt find similar result with my manual.
Thank you so much for the answer
Answer
You identified the correct function.
The help on pnorm gives the list of arguments:
pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)
with the explanation for the arguments:
x, q: vector of quantiles.
mean: vector of means.
sd: vector of standard deviations.
log, log.p: logical; if TRUE, probabilities p are given as log(p).
lower.tail: logical; if TRUE (default), probabilities are P[X <= x]
otherwise, P[X > x].
Under "Value:" it says
... ‘pnorm’ gives the distribution function,
So that covers everything. If you put the correct value you want the area to the left of in for q and the correct mu and sigma values, you will get the area below it. If you want the area above, add lower.tail=FALSE.
Like so:
pnorm(81,65,8) # area to left
[1] 0.9772499
pnorm(81,65,8,lower.tail=FALSE) # area to right ... which is what you want
[1] 0.02275013
(this way is more accurate than subtracting the first thing from 1 when you get into the far upper tail)
Edit: This diagram might clarify things:
