I am new to probability theory and still have a problem understanding some things. I came across the following problem that I am unsure on how to solve. Here it is:

Suppose that in a city with $100,000$ cars a shop sells tires for cars. It has been observed that in an interval of $3$ months the percentage of cars that come to the shop for the replacement of all their tires is $0.5$%. What is the least number of tires that the shop has to order so that it can satisfy all its customers in a $3$ month interval with probability $\geq 95$%?

And here is my attempt:

Let $X$ be the random variable of cars coming at the shop in a $3$ month interval. Then $X$ is a discrete random variable following binomial distribution $B(100,000; 0.005)$. We would like to find the least $x\in\mathbb{N}$ satisfying $P(X\leq x)\geq95$%. Since (I assume) every car has $4$ tires, our answer is going to be $4x$. Now since we have a large number of cars, by the Central limit theorem, we have that $X\sim N(\mu,\sigma^2)$, where $\mu, \sigma^2$ are the expected value and the variance of $B(100,000; 0.005)$ and therefore, by calculating, it is $\mu=500, \sigma^2=475$. Now we have $P(X\leq x)=P\big{(}\displaystyle{\frac{X-500}{\sqrt{475}}\leq\frac{x-500}{\sqrt{475}}\big{)}=P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}$, where $Z\sim N(0,1)$. Now our $x$ will satisfy $\displaystyle{P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}=0.95$, therefore $\displaystyle{P\big{(}0<Z\leq\frac{x-500}{\sqrt{475}}\big{)}}=0.45$ hence (by the tables) it is $\displaystyle{\frac{x-500}{\sqrt{475}}=1.645}$, which yields $x=535.85..$ and since $x$ is supposed to be an integer we have that $x=536$. therefore the least number of tires is $2144$.

Is my solution correct? if not, what should I have used?

Comment: I know that the $x=$fractional is not correct since i specified that $x\in\mathbb{N}$, but you get the point.

  • $\begingroup$ "the percentage of cars that come to the shop for the replacement of all their tires is 0.5%" That is quite poorly worded, it says that X is a constant ( 500). I guess it meant what you understood, though. $\endgroup$ – leonbloy Jan 21 at 14:49
  • $\begingroup$ @leonbloy I know, I was confused by this as well, but this is an accurate translation of the problem... $\endgroup$ – JustDroppedIn Jan 21 at 15:03

Assuming that the shop's only tire business is to replace all four tires, your method makes sense. [Otherwise, I can't see how to work the problem.]

The CLT implies that a binomial distribution with sufficiently large $n$ is well-approximated by a normal distribution, which is what you're doing.

Binomial. I used the binomial quantile function (inverse CDF) qbinom in R to compute an answer directly from the binomial distribution. It is very nearly the same as your answer.

4*ceiling(qbinom(.95, 10^5, .005))
[1] 2148

I expect that the small discrepancy may result from rounding in finding the mean and standard deviation of the approximating normal distribution or from using printed normal tables. [Perhaps try $\sigma^2 = 497.5.]$

Normal approximation to binomial. Without rounding, I get the following result from the normal approximation. (The normal approximation with $n=10^5$ and $p = 0.005$ should be quite good, but not necessarily exactly perfect.)

n = 10^5; p = .005; mu = n*p; sg = sqrt(n*p*(1-p))
4*ceiling(qnorm(.95, mu, sg))
[1] 2148

Poisson approximation. Another reasonable approximation is to use the distribution $\mathsf{Pois}(\lambda = \mu = np).$ This approximation is quite good for large $n$ and small $p:$

4*ceiling(qpois(.95, mu))
[1] 2148

Normal approximation to Poisson. For $\lambda$ as large as $500$ the Poisson distribution is well approximated by $\mathsf{Norm}(\lambda, \sqrt{\lambda}):$

4*ceiling(qnorm(.95, mu, sqrt(mu)))
[1] 2148

The following figure shows the binomial distribution (black bars) along with its normal approximation (blue curve) and its Poisson approximation (centers of red circles). The probability to the right of the vertical dotted line is about 0.05.

enter image description here


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