91影视

A random sample of $n=64$observations is drawn from a population with $渭=20$and $蟽=16$.

According to properties of the Sampling distribution of $\stackrel{\mathbf{炉}}{\mathbf{x}}$

${\mathit{渭}}_{\mathbf{x}}\mathbf{=}\mathit{渭}$and ${\mathit{蟽}}_{\stackrel{\mathbf{炉}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{蟽}}{\sqrt{\mathbf{n}}}$

Therefore,

${渭}_{\stackrel{炉}{x}}=20$and ${蟽}_{\stackrel{炉}{x}}=\frac{16}{\sqrt{64}}$i.e.${蟽}_{\stackrel{炉}{x}}=2$

Now,

$$\begin{gathered} P(\stackrel{炉}{x}<16)=P\left(\frac{\stackrel{炉}{x}-渭}{蟽l\sqrt{n}}<\frac{16-渭}{蟽l\sqrt{n}}\right) \\ =P\left(\frac{\stackrel{炉}{x}-20}{2}<\frac{16-20}{2}\right) \\ =P(z<-2) \end{gathered}$$

Therefore, from z-score table,

$P(\stackrel{炉}{x}<16)=0.0228$

Thus, probability that $\stackrel{炉}{x}$is less than 16 is 0.0228.

According to properties of the Sampling distribution of $\stackrel{炉}{\mathit{x}}$

${\mathit{渭}}_{\stackrel{\mathbf{炉}}{\mathbf{x}}}\mathbf{=}\mathit{渭}$and${\mathit{蟽}}_{\stackrel{\mathbf{炉}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{蟽}}{\sqrt{\mathbf{n}}}$

Therefore,

${渭}_{\stackrel{炉}{x}}=20$and ${蟽}_{\stackrel{炉}{x}}=\frac{16}{\sqrt{64}}$ i.e. ${蟽}_{\stackrel{炉}{x}}=2$

Now,

$$\begin{gathered} P(\overline{x}>23)=P\left(\frac{\overline{x}-渭}{蟽l\sqrt{n}}>\frac{23-渭}{蟽l\sqrt{n}}\right) \\ =P\left(\frac{\overline{x}-20}{2}>\frac{23-20}{2}\right) \\ =P\left(z>1.5\right) \end{gathered}$$

Therefore, from z-score table,

$$\begin{gathered} P(\overline{x}>23)=1-P\left(z<1.5\right) \\ =1-0.9332 \\ =0.0668 \end{gathered}$$

I

Thus, probability that $\overline{x}$is greater than 23 is 0.0668

According to properties of the Sampling distribution of $\overline{\mathit{x}}$

${\mathit{渭}}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{渭}$and ${\mathit{蟽}}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{蟽}}{\sqrt{\mathbf{n}}}$

Therefore,

${渭}_{\overline{x}}=20$ and ${蟽}_{\overline{x}}=\frac{16}{\sqrt{64}}$ i.e. ${蟽}_{\overline{x}}=2$

Now,

$$\begin{gathered} P(\overline{x}>25)=P\left(\frac{\overline{x}-渭}{蟽l\sqrt{n}}>\frac{25-渭}{蟽l\sqrt{n}}\right) \\ =P\left(\frac{\overline{x}-20}{2}>\frac{25-20}{2}\right) \\ =P(z>2.5) \end{gathered}$$

Therefore, from z-score table,

$$\begin{gathered} P(\overline{x}>25)=1-P(z<2.5) \\ =1-0.9937 \\ =0.0062 \end{gathered}$$

Thus, probability that $\overline{x}$is greater than 15 is 0.0062.

d.

According to properties of the Sampling distribution of $\overline{\mathit{x}}$

${\mathit{渭}}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{渭}$and ${\mathit{蟽}}_{\overline{\mathit{x}}}\mathit{=}\frac{\mathit{蟽}}{\sqrt{\mathit{n}}}$

Therefore,

${渭}_{\overline{x}=20}$ and ${蟽}_{\overline{x}}=\frac{16}{\sqrt{64}}$ i.e. ${蟽}_{\overline{x}}=2$

Now,

$$\begin{gathered} P(16<\overline{x}<22)=P\left(\frac{16-渭}{蟽l\sqrt{n}}<\frac{\overline{x}-渭}{蟽l\sqrt{n}}<\frac{22-渭}{蟽l\sqrt{n}}\right) \\ =P\left(\frac{16-20}{2}<\frac{\overline{x}-20}{2}<\frac{22-20}{2}\right) \\ =P\left(-2<z<1\right) \end{gathered}$$

Therefore, from z-score table,

$$\begin{gathered} P\left(16<\overline{x}<22\right)=P\left(-2<z<1\right) \\ =P\left(z<1\right)-P\left(z<-2\right) \\ =0.8413-0.02275 \\ P(16<\overline{x}<22)=0.8185 \end{gathered}$$

Thus, probability that $\overline{x}$falls between 16 and 22 is 0.8185.

According to properties of the Sampling distribution of $\overline{\mathit{x}}$

${\mathit{渭}}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{渭}$and ${\mathit{蟽}}_{\overline{\mathbf{x}}}\frac{\mathbf{蟽}}{\sqrt{\mathbf{n}}}$

Therefore,

${渭}_{\overline{x}}=20$and ${蟽}_{\overline{x}}=\frac{16}{\sqrt{64}}$i.e. ${蟽}_{\overline{x}}=2$

Now,

$$\begin{gathered} P\left(\overline{x}<14\right)=P\left(\frac{\overline{x}-渭}{蟽l\sqrt{n}}<\frac{14-渭}{蟽l\sqrt{n}}\right) \\ =P\left(\frac{\overline{x}-20}{2}<\frac{14-20}{2}\right) \\ =P(z<-3) \end{gathered}$$

Therefore, from z-score table,

$$\begin{gathered} P(\overline{x}<14)=P(z<-3) \\ =0.00135 \end{gathered}$$

Thus, probability that $\overline{x}$is less than 14 is 0.00135.