# Consider the marks of all 1st-year students on a statistics test. If the marks have a normal distribution with a mean of 72 and a standard deviation of 9, then the probability that a random sample of 10 students from this group have a sample mean between 71 and 73 is?

Consider the marks of all 1st-year students on a statistics test. If the marks have a normal distribution with a mean of 72 and a standard deviation of 9, then the probability that a random sample of 10 students from this group have a sample mean between 71 and 73 is?
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Step 1
Suppose, X be the marks of statistics students, X follows Normal distribution with mean 72 and standard deviation 9.
The average marks of 10 students,
$\stackrel{―}{X}=\frac{1}{10}\sum _{i=1}^{10}{X}_{i}$,
$\stackrel{―}{X}$ follows Normal distributions with mean 72 and standard deviation $\frac{9}{\sqrt{10}}$.
Step 2
The probability that the sample mean between 71 and 73 is
$P\left(71\le \stackrel{―}{X}\le 73\right)$
$=P\left(\frac{71-72}{9}/\left(\sqrt{10}\right)\le \left(\frac{\stackrel{―}{X}-72}{9}/\left(\sqrt{10}\right)\le \left(\frac{73-72}{9}/\left(\sqrt{10}\right)\right),Z=\left(\frac{\stackrel{―}{X}-72}{9}/\left(\sqrt{10}\right)\sim N\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}mal\left(0,1\right)$
$=P\left(-0.3514\le Z\le 0.3514\right)$
$=P\left(Z\le 0.3514\right)-P\left(Z\le -0.3514\right)$
$=\mathrm{\Phi }\left(0.3514\right)-\mathrm{\Phi }\left(0.3514\right),\mathrm{\Phi }\left(z\right)=P\left(Z\le z\right)$
$=2\mathrm{\Phi }\left(0.3514\right)-1,\mathrm{\Phi }\left(z\right)=1-\mathrm{\Phi }\left(-z\right)$
$=\left(2×0.6374\right)-1$
=0.2748
The values of $\mathrm{\Phi }\left(0.3514\right)$ is taken from Normal distribution.
Therefore, the probability that a random sample of 10 students from this group have a sample mean between 71 and 73 is 0.2758.
Jeffrey Jordon