I was working on statistics using R. Before i do this using R program, i have done it manually. So here is the problem.
在使用R程序之前，我正在使用R进行统计，我已经手动完成了。这就是问题所在。
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?
研究人员对300名电视观众进行了抽样调查，要求他们对电视节目的整体质量进行打分，从0(糟糕)到100(最好)。根据结果构造了一个直方图，并注意到它是堆积状且对称的，样本均值为65，样本标准差为8。大约什么比例的评级会超过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%
我已经手工回答过了:Pr(X>81)=Pr(Z>(81-65)/8)=Pr(Z>2)=0.0227，所以比例为0.023或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
我不知道怎么用R表示?我试过使用pnorm(p=..，mean=.，sd=.)，但是我的使用手册没有发现类似的结果。非常感谢你的回答
1 个解决方案
#1
3
You identified the correct function.
你找到了正确的函数。
The help on pnorm gives the list of arguments:
pnorm的帮助给出了参数列表:
pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)
with the explanation for the arguments:
并对论点作出解释:
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
在“价值:“它说
... ‘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.
这涵盖了。如果你输入正确的值你想要q左边的面积和正确的mu和sigma值，你会得到它下面的面积。如果你想要上面的区域，添加low .tail=FALSE。
Like so:
像这样:
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)
(这种方法比从1中减去第一项更准确
Edit: This diagram might clarify things:
编辑:这个图表可能会澄清一些事情:
#1
3
You identified the correct function.
你找到了正确的函数。
The help on pnorm gives the list of arguments:
pnorm的帮助给出了参数列表:
pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)
with the explanation for the arguments:
并对论点作出解释:
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
在“价值:“它说
... ‘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.
这涵盖了。如果你输入正确的值你想要q左边的面积和正确的mu和sigma值，你会得到它下面的面积。如果你想要上面的区域，添加low .tail=FALSE。
Like so:
像这样:
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)
(这种方法比从1中减去第一项更准确
Edit: This diagram might clarify things:
编辑:这个图表可能会澄清一些事情: