91Ó°ÊÓ

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

Given:

\(\begin{array}{*{20}{c}} {{\mu _X} = 105}\\ {{\sigma _X} = 8}\\ {{\mu _Y} = 100}\\ {{\sigma _Y} = 6}\\ {{n_X} = 40}\\ {{n_Y} = 35} \end{array}\)

The central limit theorem states that if the sample size is big (\({\rm{30}}\) or more), the sample mean \({\rm{\bar x}}\)sampling distribution is approximately normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar x}}\)is nearly normal because the sample size of \({\rm{40}}\)is at least \({\rm{30}}\)

The sample mean \({\rm{\bar x}}\)sampling distribution has a mean \({\rm{\mu }}\)and a standard deviation \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{\rm{\bar x}}}}{\rm{ = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = 105}}}\\{{{\rm{\sigma }}_{{\rm{\bar x}}}}{\rm{ = }}\frac{{{{\rm{\sigma }}_{\rm{X}}}}}{{\sqrt {\rm{n}} }}{\rm{ = }}\frac{{\rm{8}}}{{\sqrt {{\rm{40}}} }}{\rm{ = }}\frac{{{\rm{2}}\sqrt {{\rm{10}}} }}{{\rm{5}}}{\rm{\gg 1}}{\rm{.2649}}}\end{array}\)

The central limit theorem states that if the sample size is big (\({\rm{30}}\) or more), the sample mean \({\rm{\bar y}}\)sampling distribution is essentially normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar y}}\)is nearly normal because the sample size of \({\rm{35}}\) is at least \({\rm{30}}\)

The sample mean \({\rm{\bar y}}\)sampling distribution has a mean \({\rm{\mu }}\)and a standard deviation \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{\rm{\bar y}}}}{\rm{ = }}{{\rm{\mu }}_{\rm{Y}}}{\rm{ = 100}}}\\{{{\rm{\sigma }}_{{\rm{\bar y}}}}{\rm{ = }}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{\sqrt {\rm{n}} }}{\rm{ = }}\frac{{\rm{6}}}{{\sqrt {{\rm{35}}} }}{\rm{ = }}\frac{{{\rm{6}}\sqrt {{\rm{35}}} }}{{{\rm{35}}}}{\rm{\gg 1}}{\rm{.0142}}}\end{array}\)

For the linear combination \({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\),the mean, variance, and standard deviation have the following properties:

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}}\\{{\rm{\sigma }}_{\rm{W}}^{\rm{2}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}\left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2\;are independent\;}}} \right)}\\{{{\rm{\sigma }}_{\rm{W}}}{\rm{ = }}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}} \left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2}}} \right.{\rm{\;are independent)\;}}}\end{array}\)

Suppose, \({\rm{\bar Xand \bar Y}}\) are independent

\({{\rm{\mu }}_{{\rm{\bar x - \bar y}}}}{\rm{ = }}{{\rm{\mu }}_{{\rm{\bar x}}}}{\rm{ - }}{{\rm{\mu }}_{{\rm{\bar y}}}}{\rm{ = 105 - 100 = 5}}\)

\({{\rm{\sigma }}_{{\rm{\bar x - \bar y}}}}{\rm{ = }}\sqrt {{{\rm{\sigma }}_{{\rm{\bar x}}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{{\rm{\sigma }}_{{\rm{\bar y}}}}} {\rm{ = }}\sqrt {{\rm{1}}{\rm{.264}}{{\rm{9}}^{\rm{2}}}{\rm{ + 1}}{\rm{.014}}{{\rm{2}}^{\rm{2}}}} {\rm{\gg 1}}{\rm{.6213}}\)

Because \({\rm{\bar Xand \bar Y}}\) are approximately normally distributed, \({\rm{\bar X - \bar Y}}\)is about normally distributed (and because we assume that they are independent).

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{1 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg - 2}}{\rm{.47}}}\\{{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ - 1 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg - 3}}{\rm{.70}}}\end{array}\)

Using the normal probability table in the appendix, which gives the probabilities to the left of \({\rm{z}}\)scores, calculate the corresponding probability.

\(\begin{array}{*{20}{c}}{{\rm{P( - 1£\bar X - \bar Y£1) = P( - 3}}{\rm{.70 < Z < - 2}}{\rm{.47) = P(Z < - 2}}{\rm{.47) - P(Z < - 3}}{\rm{.70)}}}\\{{\rm{\gg 0}}{\rm{.0068 - 0 = 0}}{\rm{.0068 = 0}}{\rm{.68\% }}}\end{array}\)

The standardized score is the value \({\rm{x}}\)divided by the standard deviation after being reduced by the mean.

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{10 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg 3}}{\rm{.08}}\)

Using the normal probability table in the appendix, which gives the probabilities to the left of \({\rm{z}}\)-scores, calculate the relevant probability.

We would be skeptical of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5}}\)because the probability is so small. if was obtained.