A random sample of$\mathit{n}\mathbf{=}\mathbf{100}$observations is selected from a population with$\mathbf{\mu }\mathbf{=}\mathbf{30}$and $\mathit{\sigma }\mathbf{=}\mathbf{16}$. Approximate the following probabilities:

a.$\mathit{P}\mathbf{=}\left(\overline{x}\ge 28\right)$

b.localid="1658061042663" $\mathit{P}\mathbf{=}\mathbf{(}\mathbf{22}\mathbf{.}\mathbf{1}\mathbf{\le }\overline{\mathbf{x}}\mathbf{\le }\mathbf{26}\mathbf{.}\mathbf{8}\mathbf{)}$

c.localid="1658061423518" $\mathit{P}\mathbf{=}\mathbf{(}\overline{\mathbf{x}}\mathbf{\le }\mathbf{28}\mathbf{.}\mathbf{2}\mathbf{)}$

d.$\mathit{P}\mathbf{=}\mathbf{(}\overline{\mathbf{x}}\mathbf{\ge }\mathbf{27}\mathbf{.}\mathbf{0}\mathbf{)}$

Sample size$n=100$,$\mu =30$ and $\sigma =16$

a.

Using Central Limit Theorem $\overline{\mathbf{x}}$is approximately normally distributed and the sampling distribution of $\overline{\mathbf{x}}$has the following mean and standard deviation:

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

Therefore,

localid="1658065968111" ${\mu }_{\overline{x}}=30$,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{16}{\sqrt{100}} \\ =1.6 \end{gathered}$$

Let,

$$\begin{gathered} P\left(\overline{x}\ge 28\right)=P\left(\frac{\overline{x}-\mu }{s/\sqrt{n}}\ge \frac{28-\mu }{s/\sqrt{n}}\right) \\ =P\left(z\ge \frac{28-30}{1.6}\right) \\ =P\left(z\ge -1.25\right) \end{gathered}$$

Therefore, from z-score table,

$P\left(\overline{x}\ge 28\right)=0.8944$

Hence, the approximate probability for $\overline{x}$greater than 28 is 0.8944.

Using Central Limit Theorem, is approximately normally distributed and the sampling distribution of has the following mean and standard deviation:

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

Therefore,

${\mu }_{\overline{x}}=30$,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{16}{\sqrt{100}} \\ =1.6 \end{gathered}$$

Let,

$$\begin{gathered} P\left(22.1\le \overline{x}\le 26.8\right)=P\left(\frac{22.1-\mu }{s/\sqrt{n}}\le \frac{\overline{x}-\mu }{s/\sqrt{n}}\le \frac{26.8-\mu }{s/\sqrt{n}}\right) \\ =P\left(\frac{22.1-30}{1.6}\le z\le \frac{26.8-30}{1.6}\right) \\ =P\left(-4.937\le z\le -2\right) \\ =P\left(z\le -2\right)-P\left(z\le -4.937\right) \end{gathered}$$

Therefore, from z-score table,

$$\begin{gathered} P\left(221\le \overline{X}\le 268\right)=0.028-0.00002 \\ =0.02278 \end{gathered}$$

Hence, the approximate probability for $\overline{x}$falls between 22.1 and 26.8 is 0.8944

Using Central Limit Theorem,$\overline{\mathbf{x}}$is approximately normally distributed and the sampling distribution of $\overline{\mathbf{x}}$has the following mean and standard deviation:

localid="1658066416402" ${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and localid="1658066432641" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

localid="1658066453701" ${\mu }_{\overline{x}}=30$,

localid="1658066474405" $$\begin{gathered} {\sigma }_{\overline{x}}=\frac{16}{\sqrt{100}} \\ =1.6 \end{gathered}$$

Let,

localid="1658066506428" $$\begin{gathered} P\left(\overline{x}\le 28.2\right)=P\left(\frac{\overline{x}-\mu }{s/\sqrt{n}}\ge \frac{28.2-\mu }{s/\sqrt{n}}\right) \\ =P\left(z\ge \frac{28.2-30}{1.6}\right) \\ =P\left(z\ge -1.125\right) \end{gathered}$$

Therefore, from z-score table,

localid="1658066563911" $P\left(\overline{x}\le 28.2\right)=0.8686$

Hence, the approximate probability for $\overline{x}$less than 28.2 is 0.8686.

Using Central Limit Theorem,$\overline{\mathbf{x}}$is approximately normally distributed and the sampling distribution of $\overline{\mathbf{x}}$has the following mean and standard deviation:

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

Therefore,

${\mu }_{\overline{x}}=30$,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{16}{\sqrt{100}} \\ =1.6 \end{gathered}$$

Let,

$$\begin{gathered} P\left(\overline{x}\ge 27\right)=P\left(\frac{\overline{x}-\mu }{s/\sqrt{n}}\ge \frac{27-\mu }{s/\sqrt{n}}\right) \\ =P\left(z\ge \frac{27-30}{1.6}\right) \\ =P\left(z\ge -1.875\right) \end{gathered}$$

Therefore, from z-score table,

$P\left(\overline{x}\le 27\right)=0.9692$

Hence, the approximate probability for $\overline{x}$greater than 27 is 0.9692.