91影视

Mean$\mathrm{渭}=0.53$

Standard deviation$\mathrm{蟽}=0.193$

Sample size $\mathrm{n}=50$

According to Central Limit Theorem, if large-sized samples are drawn from any population, with mean = $渭$and standard deviation = $蟽$, then the sampling distribution of sample means approximates a normal distribution, irrespective of the underlying distribution of population. The greater the sample size, the better the approximation (typically zero). The sample mean follows the normal distribution, with the mean of the sampling mean is:

$\begin{array}{c}{\mathrm{渭}}_{\stackrel{炉}{\mathrm{x}}}=\mathrm{渭}\\ =0.53\end{array}$

The standard deviation of sample mean or standard error:

$\begin{array}{c}{\mathrm{蟽}}_{\stackrel{炉}{\mathrm{x}}}=\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\\ =\frac{0.193}{\sqrt{50}}\\ 鈮�0.0272\end{array}$

Let

$\begin{array}{c}\mathrm{P}(\stackrel{炉}{\mathrm{x}}>0.58)=\mathrm{P}\left(\frac{\stackrel{炉}{\mathrm{x}}鈭�\mathrm{渭}}{\mathrm{蟽}/\sqrt{\mathrm{n}}}>\frac{0.58鈭�\mathrm{渭}}{\mathrm{蟽}/\sqrt{\mathrm{n}}}\right)\\ =\mathrm{P}\left(\mathrm{z}>\frac{0.58鈭�0.53}{0.193/\sqrt{50}}\right)\\ =\mathrm{P}(\mathrm{z}>1.83)\end{array}$

From the z-score table,

$\mathrm{P}(\mathrm{z}>1.83)=0.0335$

Clearly, 0.59 is closer to the mean after tensioning, and that is what we would believe. In perspective, the probability of getting this extreme mean is lower than 0.0336(obtained in the last part) in the case of "before tensioning." Following is the probability of getting this sample mean in the case of "after."

Let,

$\mathrm{渭}=0.58,\mathrm{蟽}=0.193,\mathrm{n}=50$

So,

$\begin{array}{c}\mathrm{P}(\stackrel{炉}{\mathrm{x}}>0.59)=\mathrm{P}\left(\frac{\stackrel{炉}{\mathrm{x}}鈭�\mathrm{渭}}{\mathrm{蟽}/\sqrt{\mathrm{n}}}>\frac{0.59鈭�\mathrm{渭}}{\mathrm{蟽}/\sqrt{\mathrm{n}}}\right)\\ =\mathrm{P}\left(\mathrm{z}>\frac{0.59鈭�0.58}{0.193/\sqrt{50}}\right)\\ =\mathrm{P}(\mathrm{z}>0.37)\end{array}$

From the z-score table,

$\mathrm{P}(\mathrm{z}>0.37)=0.3557$