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The number of flaws per square yard in a type of carpet material varies with mean $1.6$ flaws per square yard and standard deviation $1.2$ flaws per square yard.

The population standard deviation divided by the square root of sample size equals the standard deviation of the sampling distribution of the sample mean.

${\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}=\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}$

As a result, the standard deviation lowers as the sample size grows, and the data values become closer to the predicted value as the sample size grows.

As a result, when the sample size is bigger, you are less likely to find a mean of two or more flaws, and when the sample size is less, you are more likely to find a mean of two or more flaws.

The number of flaws per square yard in a type of carpet material varies with mean$1.6$ flaws per square yard and standard deviation $1.2$ flaws per square yard.

Consider ,$\mathrm{n}=1$

The center limit theorem states that if the sample size is more than 30, the sampling distribution of the sample mean $\stackrel{¯}{x}$is approximately normal.

The central limit theorem cannot be applied since the sample size of 1 is less than $30$. The sample mean sampling distribution has the same shape as the population distribution in this example.

Although the population distribution is unknown, the form of the sampling distribution of the sample mean is also unknown, which means that the probability cannot be calculated.

The number of flaws per square yard in a type of carpet material varies with mean $1.6$ flaws per square yard and standard deviation$1.2$flaws per square yard.

Consider that ,

$\begin{array}{r}\mathrm{μ}=16\\ \mathrm{σ}=1.2\\ \mathrm{n}=50\\ \stackrel{¯}{\mathrm{x}}=2\end{array}$

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}$
$\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}$

z-score is localid="1657630095652" $z=\frac{x-{\mathrm{μ}}_{\stackrel{¯}{x}}}{{\mathrm{σ}}_{\stackrel{¯}{x}}}=\frac{\stackrel{¯}{x}-\mathrm{μ}}{\mathrm{σ}/\sqrt{n}}=\frac{2-1.6}{1.2\sqrt{50}}=2.36$

The z-two score's found expressions must then be equal:

localid="1657629821949" $\mathrm{P}(\mathrm{Z}<2.36)$is given in the row starting with$.01$and in the column starting with $.06$of the standard normal probability table

localid="1657629877918" $\begin{array}{c}\mathrm{P}(\stackrel{¯}{\mathrm{X}}â‰\yen 2)=\mathrm{P}(\mathrm{Z}>2.36)\\ =1−\mathrm{P}(\mathrm{Z}<2.36)\\ =1−0.9209\\ =0.0081\\ =0.81\mathrm{\%}\end{array}$