91影视

Random sample varies normal around the target mean $渭$ with standard deviation $蟽$.

Rule of Multiplication:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

Rule of Complement

$P(notA)=1-P\left(A\right)$

The $68-95-99.7$-rule informs us that $95\%$ of the mean of the sample is within $2蟽$ of the mean and therefore $5\%$ of the mean of the samples is greater than $2蟽$ of the mean.

$P(within2蟽fromthemean)=95\%=0.95$

$\mathrm{P}\left(\text{greater than}2蟽\text{from the mean}\right)=5\%=0.05$

Rule of Multiplication:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

Suppose that $\mathrm{X}$is the number of samples, which implies that the mean is more than $P(X=0)=P(\text{within}2蟽\text{from the mean}{)}^2$

$=0.{95}^2$

$=0.9025$

Rule of Complement

$P(notA)=1-P\left(A\right)$

It could then determine the chances of having more than $2蟽$from the mean of minimum one of the two samples.

$P(X鈮�1)=1-P(X=0)$

$=1-0.9025$

$=0.0975$

Random sample varies normal around the target mean $渭$with standard deviation $蟽$.

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

The $68-95-99.7$-rule informs us that $95\%$of the mean of the sample is within $2蟽$of the mean and therefore $5\%$of the mean of the samples is greater than $2蟽$of the mean.

$P(\text{larger than}渭+2蟽)=2.5\%=0.025$$P(\text{less than}渭+2蟽)=97.5\%=0.975$

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

If the $4^{\text{th}}$sample is the $1^{\text{st}}$sample with a mean larger than $渭+2蟽$,

Then the $3$previous samples have a mean less than $渭+2蟽$.

localid="1650392399658" $P(1\text{st larger than}渭+2蟽\text{on 4th sample})=P(\text{less than}渭+2蟽{)}^3脳P(\text{larger than}渭+2蟽)$

localid="1650392417061" $$\begin{gathered} =0.{975}^3脳0.025 \\ =0.0232 \end{gathered}$$

Random sample varies normal around the target mean $渭$ with standard deviation $蟽$.

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

The $68-95-99.7$-rule informs us that $95\%$of the mean of the sample is within $2蟽$of the mean and therefore $5\%$of the mean of the samples is greater than $2蟽$of the mean.

Sample mean in $(渭-蟽,渭+蟽)$.

Probability $=68\%=0.68$.

Sample mean not in $(渭-蟽,渭+蟽)$.

Probability $=32\%=0.32$.

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)脳P\left(B\right)$

Suppose $\mathrm{X}$is the number of samples means in $(渭-蟽,渭+蟽)$from the$5$-sample means.

Sample mean not in $(渭-蟽,渭+蟽)$

$P(X=0)=P\left(\text{Sample mean not in}\right(渭-蟽,渭+蟽){)}^5=0.{32}^5=0.003355$

Sample mean not in $(渭-蟽,渭+蟽)$.

$P(X=0)=5脳P\left(\text{Sample mean not in}\right(渭-蟽,渭+蟽){)}^4$

$脳P\left(\text{Sample mean not in}\right(渭-蟽,渭+蟽\left)\right)$

$=5脳0.{32}^4脳0.68=0.035652$

Add the associating probability

$P(X鈮�1)=0.003355+0.035652$

$=0.039007$

$=3.9007\%$

Since the probability is below $5$percent, it is unlikely that this occurrence will happen by chance and this is thus a reasonable criterion.