91影视

It is given that random samples vary normally around the target mean with a standard deviation.

The$68-95-99.7$tells us that$95\%$of the sample is within$2$standard deviation of the mean and thus$5\%$of the samples mean is more than$2$standard deviation of the mean.

By multiplication rule we know that,

$P(AandB)=P\left(A\right)脳P\left(B\right)$

Let X be the number of sample means that are more than$2$standard deviation of the mean.

Thus,

$P(X=0)=P{(within2s\tan darddeviationfromthemean)}^2=0.{95}^2=0.9025$

Now, by the complement rule we have,

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

We can then determine the probability of having at least one of the two sample means more than standard deviation of the mean. This is,

$$\begin{gathered} P(X鈮�1)=1-P(X=0) \\ =1-0.9025 \\ =0.0925 \end{gathered}$$

Since the normal distribution is symmetric about its mean, $2.5\%$of the sample mean is greater than $2蟽+渭$

By multiplication rule we know that,

$P(AandB)=P\left(A\right)脳P\left(B\right)$

If the fourth sample is the first sample with a mean greater than $2蟽+渭$, then the three previous sample had a mean less than $2蟽+渭$

$$\begin{gathered} P(firstgreaterthan2蟽+渭onfourthsample) \\ =P{(lessthan2蟽+渭)}^3脳P(greaterthan2蟽+渭) \\ =0.{975}^3脳0.025 \\ =0.0232 \end{gathered}$$

By multiplication rule we know that,

$P(AandB)=P\left(A\right)脳P\left(B\right)$

Let X be the number of sample means in $(渭+蟽,渭-蟽)$out of the five sample means.

$$\begin{gathered} P(X=0)=P(Samplemeannotin{(渭+蟽,渭-蟽))}^5=0.{32}^5=0.003355 \\ P(X=1)=5脳P(Samplemeannotin{(渭+蟽,渭-蟽))}^4脳P(Samplemeanin(渭+蟽,渭-蟽\left)\right) \\ =5脳0.{32}^4脳0.68 \\ =0.035652 \end{gathered}$$

Thus, we have,

$$\begin{gathered} P(X鈮�1)=0.003355+0.035652 \\ =0.039007 \\ =3.9007\% \end{gathered}$$

Since the probability is less than$5\%$ this event is unlikely to happen by chance and thus is a reasonable criteria.