Let $X_1,X_2,…,X_{100}$ be a random sample from a distribution with pdf $f(x) = \cfrac{4}{3} – x^2, 0 \leq x \leq 1$

Question

Use the central limit theorem to find an approximate probability of $P(0.4 < \bar X < 0.5)$

So I've found the mean ($\mu$) to be $\cfrac{5}{12}$

and I've found the Variance ($\sigma^2$) to be $\cfrac{17}{240}$

So standard deviation ($\sigma) $ is $ .2661$ (approximately)
So my thought process was
$P(0.4 < \bar X < 0.5)$ = $P(\cfrac{0.4-\cfrac{5}{12}}{.2661/10} < \bar X < \cfrac{0.5-\cfrac{5}{12}}{.2661/10})$ = $P(-.63 < Z < 3.13) = .7348$

But my book says that the answer should be $.9342$

What on earth am I doing wrong here? Any help would be greatly appreciated.

Answer 1

I agree with your calculations.

As a further check, here is a simulation in R:

set.seed(2020)
cases <- 10^7
samplesize <- 100
densitytest <- function(x,t){ return(t*4/3 < 4/3 - x^2) }
dat <- runif(cases)
testdat <- runif(cases)
okdat <- dat[densitytest(dat,testdat)]
okdat <- okdat[1:(samplesize * floor(length(okdat)/samplesize))]
mean(okdat)
# 0.4166633
sd(okdat) 
# 0.2661291
matdat <- matrix(okdat, ncol=samplesize)
meandat <- rowMeans(matdat)
mean(meandat)
# 0.4166633
sd(meandat)
# 0.02664449
mean(0.4 < meandat & meandat < 0.5)
# 0.7317877

which is close to (though slightly lower than) your normal approximation