Question

A tobacco company claims that the amount of nicotine in one of its cigarettes is a random variable with a mean of $2.2$mg and a standard deviation of $0.3$mg. However, the average nicotine content of $100$randomly chosen cigarettes was $3.1$mg. Approximate the probability that the average would have been as high as or higher than $3.1$ if the company’s claims were true

Short answer

If the company claims were true, the probability is approximately equal to$0.$

Step-by-step solution

Given information

The random variable indicating the amount of nicotine has a mean of $2.2mg$and standard deviation of $.3mg$

The average nicotine content of $100$chosen cigarettes was $3.1mg$

Approximate the probability.

Explanation

Let $X_i$represents the amount$\left(inmg\right)$of nicotine in $i$th cigarette. It is given that $X_i$is a random variable with a mean that a tobacco company claims

$\mu ={\mu }_i=E\left[X_i\right]=2.2$

And the standard deviation is

$\sigma ={\sigma }_i=\sqrt{Var\left(X_i\right)}=.3$

Let $\overline{X}$represents the average nicotine content of $100$randomly chosen cigarettes.

Without sacrificing generality,

$\overline{X}=\frac{X_1+X_2+\cdots +X_{100}}{100}$

and since $X_1,X_2,\dots$is a set of randomly distributed random variables that are all independent and have the same mean $\mu$and variance $\sigma$, by The central limit theorem, for each $a\in \mathrm{ℝ}$

$P\left\{\frac{\frac{X_1+X_2+\cdots +X_n}{n}-\mu }{\frac{\sigma }{\sqrt{n}}}\le a\right\}\to \Phi \left(a\right),n\to \infty .$

Final answer

By the given terms, if the company claims were true

$$\begin{gathered} P\{\overline{X}\ge 3.1\}=1-P\{\overline{X}\le 3.1\}=1-P\left\{\frac{\overline{X}-\mu }{\frac{\sigma }{10}}\le \frac{3.1-\mu }{\frac{\sigma }{10}}\right\} \\ \approx 1-\Phi \left(\frac{3.1-\mu }{\frac{\sigma }{10}}\right)=1-\Phi \left(\frac{3.1-2.2}{\frac{.3}{10}}\right) \\ =1-\Phi \left(30\right)=1-1=0 \end{gathered}$$

Therefore, the probability is approximately equal to zero.