91影视

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.

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

$渭={渭}_i=E\left[X_i\right]=2.2$

And the standard deviation is

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

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

Without sacrificing generality,

$\stackrel{炉}{X}=\frac{X_1+X_2+鈰�+X_{100}}{100}$

and since $X_1,X_2,鈥�$is a set of randomly distributed random variables that are all independent and have the same mean $渭$and variance $蟽$, by The central limit theorem, for each $a鈭�\mathrm{鈩�}$

$P\left\{\frac{\frac{X_1+X_2+鈰�+X_n}{n}-渭}{\frac{蟽}{\sqrt{n}}}鈮�a\right\}鈫�桅\left(a\right),n鈫�鈭�.$

By the given terms, if the company claims were true

$$\begin{gathered} P\{\stackrel{炉}{X}鈮�3.1\}=1-P\{\stackrel{炉}{X}鈮�3.1\}=1-P\left\{\frac{\stackrel{炉}{X}-渭}{\frac{蟽}{10}}鈮�\frac{3.1-渭}{\frac{蟽}{10}}\right\} \\ 鈮�1-桅\left(\frac{3.1-渭}{\frac{蟽}{10}}\right)=1-桅\left(\frac{3.1-2.2}{\frac{.3}{10}}\right) \\ =1-桅\left(30\right)=1-1=0 \end{gathered}$$

Therefore, the probability is approximately equal to zero.