91Ó°ÊÓ

The expectation of a random variable's squared deviation from its population mean or sample mean is called variance in probability theory and statistics. Variance is a measure of dispersion, or how far a set of numbers deviates from their average value.

Given:

\(\begin{array}{l}\mu = 5.00\\\sigma = 0.2\\n = 25\end{array}\)

(a)If X is normally distributed with mean\(\mu \)and standard deviation\(\sigma \), then the sample mean\(\bar X\)is also normally distributed with mean\(\mu \)and standard deviation\(\frac{\sigma }{{\sqrt n }}\). (Same for\(\bar Y\))

\(\begin{array}{l}{\mu _{\bar X}} = {\mu _{\bar Y}} = \mu = 5.00\\{\sigma _{\bar X}} = {\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }} = \frac{{0.2}}{{\sqrt {25} }} = \frac{{0.2}}{5} = 0.04\end{array}\)

If\(\bar X\)and \(\bar Y\)are both normally distributed, then their linear combination is also normally distributed and thus \(\bar X - \bar Y\)is also normally distributed.

For the linear combination\(W = a{X_1} + b{X_2}\), the mean, variance, and standard deviation are as follows:

\(\begin{array}{l}{\mu _W} = a{\mu _1} + b{\mu _2}\sigma _W^2 = {a^2}\sigma _1^2 + {b^2}\sigma _2^2{\rm{ (If }}{X_ - }1{\rm{ and }}{X_ - }2{\rm{ are independent) }}\\{\sigma _W} = \sqrt {{a^2}\sigma _1^2 + {b^2}\sigma _2^2} {\rm{ (If }}{X_ - }1{\rm{ and }}{X_ - }2{\rm{ are independent) }}\end{array}\)

For\(W = \bar X - \bar Y\)(with\(a = 1\)and\(b = - 1\)), we then obtain:

\({\mu _{\bar X - \bar Y}} = {\mu _{\bar X}} - {\mu _{\bar Y}} = 5.00 - 5.00 = 0\)

\({\sigma _{\bar X - \bar Y}} = \sqrt {\sigma _{\bar X}^2 + \sigma _{\bar Y}^2} = \sqrt {{{0.04}^2} + {{0.04}^2}} = \sqrt {0.0032} \approx 0.0565685 \ldots \)

The standardized score is the value x decreased by the mean and then divided by the standard deviation.

\(\begin{array}{l}z = \frac{{x - \mu }}{\sigma } = \frac{{ - 0.1 - 0}}{{\sqrt {0.0032} }} \approx - 1.77\\z = \frac{{x - \mu }}{\sigma } = \frac{{0.1 - 0}}{{\sqrt {0.0032} }} \approx 1.77\end{array}\)

Determine the corresponding probability using table.

\(\begin{array}{c}P( - 0.1 \le \bar X - \bar Y \le 0.1) = P( - 1.77 < Z < 1.77)\\ = P(Z < 1.77) - P(Z < - 1.77)\\ = 0.9616 - 0.0384\\ = 0.9232\\ = 92.32\% \end{array}\)

(b)

\(n = 36\)

If X is normally distributed with mean\(\mu \)and standard deviation\(\sigma \), then the sample mean\(\bar X\)is also normally distributed with mean\(\mu \)and standard deviation\(\frac{\sigma }{{\sqrt n }}\). (Same for\(\bar Y\))

\(\begin{array}{l}{\mu _{\bar X}} = {\mu _{\bar Y}} = \mu = 5.00\\{\sigma _{\bar X}} = {\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }} = \frac{{0.2}}{{\sqrt {36} }} = \frac{{0.2}}{6} = \frac{1}{{30}} \approx 0.0333\end{array}\)

If \(\bar X\)and \(\bar Y\)are both normally distributed, then their linear combination is also normally distributed and thus \(\bar X - \bar Y\)is also normally distributed.

For the linear combination\(W = a{X_1} + b{X_2}\), the mean, variance, and standard deviation are as follows:

\(\begin{array}{l}{\mu _W} = a{\mu _1} + b{\mu _2}\sigma _W^2 = {a^2}\sigma _1^2 + {b^2}\sigma _2^2{\rm{ (If }}{X_ - }1{\rm{ and }}{X_ - }2{\rm{ are independent)}}\\{\sigma _W} = \sqrt {{a^2}\sigma _1^2 + {b^2}\sigma _2^2} {\rm{ (If }}{X_ - }1{\rm{ and }}{X_ - }2{\rm{ are independent) }}\end{array}\)

For\(W = \bar X - \bar Y\)(with\(a = 1\)and\(b = - 1\)), we then obtain:

\(\begin{array}{l}{\mu _{\bar X - \bar Y}} = {\mu _{\bar X}} - {\mu _{\bar Y}} = 5.00 - 5.00 = 0\\{\sigma _{\bar X - \bar Y}} = \sqrt {\sigma _{\bar X}^2 + \sigma _{\bar Y}^2} = \sqrt {{{\left( {\frac{1}{{30}}} \right)}^2} + {{\left( {\frac{1}{{30}}} \right)}^2}} = \frac{{\sqrt 2 }}{{30}} \approx 0.0471\end{array}\)

The standardized score is the value x decreased by the mean and then divided by the standard deviation.

\(\begin{array}{l}z = \frac{{x - \mu }}{\sigma } = \frac{{ - 0.1 - 0}}{{\sqrt 2 /30}} \approx - 2.12\\z = \frac{{x - \mu }}{\sigma } = \frac{{0.1 - 0}}{{\sqrt 2 /30}} \approx 2.12\end{array}\)

Determine the corresponding probability using table.

\(\begin{array}{c}P( - 0.1 \le \bar X - \bar Y \le 0.1) = P( - 2.12 < Z < 2.12)\\ = P(Z < 2.12) - P(Z < - 2.12)\\ = 0.9830 - 0.0170\\ = 0.9660\\ = 96.60\% \end{array}\)