91Ó°ÊÓ

$\begin{array}{r}\mathrm{μ}=9.5\\ \mathrm{σ}=1.0\\ \mathrm{n}=5\end{array}$

The following formula was used:

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

The sampling distribution of the sample mean is normal because the population distribution is normal $\stackrel{¯}{\mathrm{x}}$This is also typical.

The sample mean has a mean of the sampling distribution.

${\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}=\mathrm{μ}=9.5$

The standard deviation of the sampling distribution of the sample mean is

${\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}=\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}=\frac{1.0}{\sqrt{5}}=0.4472$

As a result, the sample mean's sampling distribution is Normal with mean $9.5$and standard deviation$0.4472$

$\begin{array}{r}\mathrm{μ}=9.5\\ \mathrm{σ}=1.0\\ \mathrm{n}=5\\ \mathrm{x}=9\text{or}10\end{array}$

The following formula was used:

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

The sampling distribution of the sample mean is normal because the population distribution is normal $\stackrel{¯}{\mathrm{x}}$is also typical.

The Z-score is

$\begin{array}{l}\mathbf{z}\mathbf{=}\frac{\mathbf{x}\mathbf{−}{\mathbf{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}}{{\mathbf{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}}\mathbf{=}\frac{\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{−}\mathbf{μ}}{\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}}\mathbf{=}\frac{\mathbf{9}\mathbf{−}\mathbf{9}\mathbf{.5}}{\frac{\mathbf{1}\mathbf{.01}}{\sqrt{\mathbf{5}}}}\mathbf{=}\mathbf{−}\mathbf{1}\mathbf{.12}\\ \mathbf{z}\mathbf{=}\frac{\mathbf{x}\mathbf{−}{\mathbf{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}}{{\mathbf{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}}\mathbf{=}\frac{\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{−}\mathbf{μ}}{\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}}\mathbf{=}\frac{\mathbf{191}\mathbf{−}\mathbf{188}}{\frac{\mathbf{41}}{\sqrt{\mathbf{100}}}}\mathbf{=}\mathbf{−}\mathbf{0}\mathbf{.7}\end{array}$

The normal probability is used to calculate the associating probability.

$P(Z<−1.12)$is given in the first row, beginning with $−1.1$in the column that begins with 0.2 of the normal probability distribution $\mathrm{P}(\mathrm{Z}<1.12)$is given in the first row, beginning with 1.1 in the column that begins with 0.2 of the normal probability distribution

$\begin{array}{c}\mathrm{P}(9<\stackrel{¯}{\mathrm{X}}<10)=\mathrm{P}(−1.12<\mathrm{Z}<1.12)\\ =\mathrm{P}(\mathrm{Z}<1.12)−\mathrm{P}(\mathrm{Z}<−1.12)\\ =0.8686−0.1314=0.7372=73.72\end{array}$

$\begin{array}{r}\mathrm{μ}=9.5\\ \mathrm{σ}=1.0\\ \mathrm{n}=5\\ \mathrm{x}=9\text{or}10\end{array}$

The following formula was used:

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

The sampling distribution of the sample mean is normal because the population distribution is normal $\stackrel{¯}{\mathrm{x}}$is also typical.

Z-score is

$\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}=\frac{\stackrel{¯}{\mathrm{x}}−\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}}=\frac{9−9.5}{\frac{1.0}{\sqrt{50}}}=−3.54$

The normal probability is used to calculate the associating probabilitylocalid="1654674407714" $\mathrm{P}(\mathrm{Z}<-3.54)$is given in the first row, beginning with -3.5 in the column that begins with .04, In the row beginning with, the normal probability table is shown in its most basic form 3.5 and in the column that begins with .04 of the normal probability distribution

$\begin{array}{l}\mathrm{P}(9<\stackrel{¯}{\mathrm{X}}<10)=\mathrm{P}(−3.54<\mathrm{Z}<3.54)\\ =\mathrm{P}(\mathrm{Z}<3.54)−\mathrm{P}(\mathrm{Z}<−3.54)\\ =0.9998−0.0002\\ =0.9996\\ =99.96\%\end{array}$

The larger sample is "better" because the probability of the sample mean being within 0.5 of the population mean is higher with a larger sample, hence our estimations of the population mean are more accurate.