Refer to Exercise 5.18. Find the probability that

1. $\overline{x}$is less than 16.
2. $\overline{x}$is greater than 23.
3. $\overline{x}$is greater than 25.
4. $\overline{x}$falls between 16 and 22.
5. $\overline{x}$ is less than 14.

A random sample of $n=64$observations is drawn from a population with $\mu =20$and $\sigma =16$.

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

${\mathit{\mu }}_{\mathbf{x}}\mathbf{=}\mathit{\mu }$and ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

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

Now,

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

Therefore, from z-score table,

$P(\overline{x}<16)=0.0228$

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

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

${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

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

Now,

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

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 $\overline{\mathit{x}}$

${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

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

Now,

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

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 $\overline{\mathit{x}}$

${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and ${\mathit{\sigma }}_{\overline{\mathit{x}}}\mathit{=}\frac{\mathit{\sigma }}{\sqrt{\mathit{n}}}$

Therefore,

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

Now,

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

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 $\overline{\mathit{x}}$

${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

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

Now,

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

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.