91Ó°ÊÓ

If a random variable X from the population follows \(X \sim N\left( {\mu ,\sigma } \right)\), then the sample mean follows \(\overline x \sim N\left( {\mu ,\sqrt {\frac{{{\sigma ^2}}}{n}} } \right)\)

Sample 1: A random sample of16 observations is drawn from the normal distribution with mean \(\mu \) and standard deviation 12.

Therefore,

n=16

\(\overline x \sim N\left( {\mu ,\sqrt {\frac{{{{12}^2}}}{{16}}} } \right) \ldots \left( 1 \right)\)

Sample 2: A random sample of25 observations is drawn from the normal distribution with the same mean \(\mu \) and standard deviation of 20.

Therefore,

n=25

\(\overline y \sim N\left( {\mu ,\sqrt {\frac{{{{20}^2}}}{{25}}} } \right) \ldots \left( 2 \right)\)

Let,

\(W = \overline X - \overline Y \)

Then the expectation and variance are

\(\begin{aligned}{}E\left( W \right) &= E\left( {\overline X - \overline Y } \right)\\ &= E\left( {\overline X } \right) - E\left( {\overline Y } \right)\\ &= \mu - \mu\\ &= 0\end{aligned}\)

\(\begin{aligned}{}V\left( W \right) &= V\left( {\overline X - \overline Y } \right)\\ &= V\left( {\overline X } \right) + V\left( {\overline Y } \right)\\ &= \frac{{{{12}^2}}}{{16}} + \frac{{{{20}^2}}}{{25}}\,\,,\,\,from\,\,\left( 1 \right)\,\,and\,\,\left( 2 \right)\\ &= 25\end{aligned}\)

Since the samples are drawn independently, their covariance term is 0.

Therefore,

\(W \sim N\left( {0,5} \right)\)

\(\begin{aligned}{} &= \Pr \left( {\left| {\overline x - \overline y } \right| < 5} \right)\\& = \Pr \left( {\left| w \right| < 5} \right)\\ &= \Pr \left( {\frac{{w - 0}}{5} < \frac{{5 - 0}}{5}} \right)\\ &= \Pr \left( {z < 1} \right)\\ &= 0.842\end{aligned}\)

Therefore, the answer is 0.842