Find approximation to the probability that the total mass of 64 randomly selected tourists, is more than 4200kg.

Question

A tourist bridge can only support the mass of 4200kg. We don't know the pdf of the mass ($X$, in kg) of a single tourist. but we do know that $E(X)=65$, $Var(X)=100$. find good approximation to the probability that the total mass of 64 randomly selected tourists, is more than 4200kg.

MY working:

As far as I understand this statement seems to imply the use of central limit theorem. May be I am wrong, and if I am right I am unable to use central limit theorem to answer question. Please guide me .

Answer 1

Comment continued. It is almost always useful to make a rough sketch of the applicable normal distribution or of the standard normal distribution. In each plot below (made using R) the desired probability is represented by the area under the density curve to the right of the vertical dotted line. The total area under each density curve is $1.$

par(mfrow=c(1,2))
 hdr1 = "Density of NORM(4160, 80)"
 curve(dnorm(x, mu, sg), 3900, 4500, ylab="Density", 
       xlab="y", lwd=2, main=hdr1)
  abline(h=0, col="green2")
  abline(v=4200, col="red", lwd=2, lty="dotted")
 hdr2 = "Density of NORM(0, 1)"
 curve(dnorm(x), -3, 3, ylab="Density", 
       xlab="z", lwd=2, main=hdr2)
  abline(h=0, col="green2")
  abline(v=0, col="green2")
  abline(v=0.5, col="red", lwd=2, lty="dotted")
par(mfrow=c(1,1))

Guidelines for making rough sketches: (a) The peak of the normal curve is at $\mu.$ (b) The curve almost touches the horizontal axis at $\pm 3\sigma.$ (c) The curvature in the center is convex and the curvature in the tails is concave; the type of curvature changes at $\mu \pm \sigma.$

Answer 2

In the absence of any other information, CLT is the way to go here. The weight of $N$ tourists added together (denote it $Y_N$) can be approximated by a normal RV with mean $N \times 65$ and standard deviation $\sqrt{N} \times 10$.

You have to compute $\mathbb{P}(Y_N \geq 4200)$. It is a good idea to write $Y_N = 10\sqrt{N} \times Z + 65N$, where $Z$ is a normal RV with zero mean and unit variance. You can check that $\mathbb{E}[Y_N] = 65N$ and $\mathbb{\sigma}[Y_N] = 10\sqrt{N}$.

$\mathbb{P}(Y_N \geq 4200) = \mathbb{P}(10\sqrt{N} \times Z + 65N \geq 4200)$

$= \mathbb{P}(Z \geq \frac{4200 - 65N}{10N})$

$=Q\left( \frac{4200 - 65N}{10\sqrt{N}} \right)$

where $Q$ is the complimentary CDF of a zero mean, unit variance normal RV. It's built into pretty much every computational library (either as Q-function or it's cousin, erfc)

Setting $N=64$, you get the probability as 30.85%.