The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation of 0.03 grams.
mean = 2.5
sd = 0.03
#using the Z score formula
(2.4-mean)/sd
## [1] -3.333333
#we can assume ND so we can use pnorm to get the prob.
pnorm(2.4, mean = mean, sd = sd)
## [1] 0.0004290603
There is a 0.04% chance of a penny weighing less than 2.4 grams.
since we are now sampling from a nearly normal distribution, this means we are sampling from a sample distribution.
This means that our standard of error for the new sample is the population sigma divided by the square root of the sample size.
n=10
SDsample = sd/n^0.5
SDsample
## [1] 0.009486833
This means that we are also making the assumption that the sample of 10 pennies is also approximately normal with mean = the sample population mean and standard error as sigma / sqrt n N(mean, sd/n^0.5) or 0.009
in this case the Z score still uses the population mean but now uses the sigma of the population which is sd/sqrt n
(2.4-2.5)/SDsample
## [1] -10.54093
pnorm(2.4, mean = mean, sd = SDsample)
## [1] 2.797279e-26
the probability is so small it is virtually 0.
#plot 1 sample population
x <- seq(2.4,2.6, length=100)
y1 <- dnorm(x=x, mean=mean, sd=sd)
#plot 2 sample from population
y2 <- dnorm(x=x, mean=mean, sd = SDsample)
#plotted together
par(mfrow=c(1,1))
plot(x,y1, main ="N(mu = mean, sigma = sd)")+
plot(x,y2, main ="N(mu = mean, sigma = SD Sample, n=10)")
## numeric(0)
For a, we wouldn’t be able to since the population itself is skewed and couldn’t be normal. However for c, if it met the conditions for being nearly normal (IID, n>=30, and not STRONGLY skewed) then yes we would be able to estimate probbilities, however at a sample of n=10 it is not possible.