Question

Refer to Exercise 5.18. Find the probability that

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

Short answer

1. Probability that $\overline{x}$is less than 16 is 0.0228.

1. Probability that $\overline{x}$is greater than 23 is 0.0668.

1. Probability that $\overline{x}$is greater than 15 is 0.0062.

1. Probability that $\overline{x}$falls between 16 and 22 is 0.8185.

1. Probability that $\overline{x}$is less than 14 is 0.00135.

Step-by-step solution

Given information

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

Computing the probability that x¯ is less than 16 

According to properties of the Sampling distribution of $\overline{\mathbf{x}}\mathbf{}{\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$

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,

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

Therefore, from z-score table,

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

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

Computing the probability that x¯ is greater than 23 

1. $\overline{x}$

According to properties of the Sampling distribution of localid="1660821731616" $\overline{\mathbf{x}}\mathbf{}{\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$

andlocalid="1660821735110" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,${\mu }_{\overline{x}}=20$

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

Now,

localid="1658202322115" $$\begin{gathered} P(\overline{x}>23)=P\left(\frac{\overline{x}-\mu }{\sigma l\sqrt{n}}>\frac{23-\mu }{\sigma l\sqrt{n}}\right) \\ =P\left(\frac{\overline{x}-20}{2}>\frac{23-20}{2}\right) \\ =P(z>1.5) \end{gathered}$$

Therefore, from z-score table,

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

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

Computing the probability that x¯ is greater than 25 

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

localid="1660821743033" ${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$andlocalid="1660821747012" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

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

Now,

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

Therefore, from z-score table,

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

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

Computing the probability that  falls between 16 and 22 

According to properties of the Sampling distribution of $\overline{\mathbf{x}}\mathbf{}\mathbf{}\mathbf{}{\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$

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,

Therefore, from z-score table,

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

Computing the probability that is x¯ less than 14 

According to properties of the Sampling distribution of $\overline{\mathbf{x}}\mathbf{}\mathbf{}\mathbf{}{\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$

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,

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

Therefore, from z-score table,

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

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