Question

:A random sample of n = 68 observations is selected from a population with$\mu =19.6$and $\sigma =3.2$Approximate each of the following probabilities

a)$p\left(\overline{X}⩽19.6\right)$

b)$p\left(\overline{X}⩽19\right)$

c)$p\left(\overline{X}⩾20.1\right)$

d)$p\left(19.2⩽\overline{X}⩽20.6\right)$

Short answer

a)

b)

c)

d)

Step-by-step solution

Given Information

The number of sample size is 68.

The mean and standard deviation are 19.6 and 3.2

(a) Compute the probability for given condition

$$\begin{gathered} \left(\overline{X}⩽19.6\right)=p\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{19.6-19.6}{\frac{3.2}{\sqrt{68}}}\right) \\ =p\left(z⩽\frac{0}{0.388}\right) \\ =0.5 \end{gathered}$$

Therefore,$p\left(\overline{X}⩽19.6\right)=0.5$

(b) Calculation

$$\begin{gathered} \left(\overline{X}⩽19\right)=p\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{19-19.6}{\frac{3.2}{\sqrt{68}}}\right) \\ =p\left(z⩽\frac{-0.6}{0.388}\right) \\ =0.061 \end{gathered}$$

Therefore$p\left(\overline{X}⩽19\right)=0.061$

(c) Calculation

$$\begin{gathered} \left(\overline{X}⩾20.1\right)=p\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{20.1-19.6}{\frac{3.2}{\sqrt{68}}}\right) \\ =p\left(z⩽\frac{0.5}{0.388}\right) \\ =0.098 \end{gathered}$$

$p\left(\overline{X}⩽19\right)=0.061$

(d) Calculation

$$\begin{gathered} p\left(19.2⩽\overline{X}⩽20.6\right)=p\left(\frac{19.2-19.6}{\frac{3.2}{\sqrt{68}}}⩽\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{20.6-19.6}{\frac{3.2}{\sqrt{68}}}\right) \\ =p\left(\frac{-0.4}{0.388}⩽z⩽\frac{1}{0.388}\right) \\ =p\left(z⩽\frac{1}{0.388}\right)-p\left(z⩽\frac{-0.4}{0.388}\right) \\ =0.99-0.15 \\ =0.84 \end{gathered}$$

Therefore,$p\left(19.2⩽\overline{X}⩽20.6\right)=0.84$