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**Answer**

## Examples of Normal Distribution

**A random variable x has a normal distribution with mean 9 and a standard deviation of 3. Find the probability P(5<x<11)**

When x=5

$z=\frac{5-9}{3}=-1.333$

When x=11

$z=\frac{11-9}{3}=0.66667$

pr(-1.333<z<0.66667)

pr(z<0.66667)

=0.7486

1-pr(z<1.333)=pr(z<-1.333)

=1-0.9082

=0.0918

pr(-1.3333<z<0.66667)

=0.7486-0.0918

=0.6568

**2. The length of a semester approximates to a normal distribution with a mean of 266 days and a standard deviation of 16 days. What proportion of the classes will last between 240 and 270 days (Tentatively between 8 and 9 months)**

The Normal Distribution will be given by;

$z=\frac{x-\mu}{\delta}$

$\mu =226andthe\delta =16$

$z=\frac{240-266}{16}=-1.625$

$z=\frac{270-266}{16}=0.25$

$p(240<x<270)=P(-1.625<z<0.25)$

$P(-1.625<z<0.25)=P(z<0.25)-p(Z<-1.63)$

$P(-1.625<z<0.25)=0.5987-0.0516$

=0.5471

**In some instances you may not be given the value of x.**

For instance, using the data in question 2, what is the length of time that marks the shortest 70% of the semester?

$U\mathrm{sin}gNormalDistribution\mu =266and\delta =16$

$P(x<?)=0.70$

This implies that;

$P(Z<?)=0.70$

Z=0.52

Thus the value of x will be;

$x=266+0.52\left(16\right)$

$X=266+8.32$

x=274.32

**Thus the time in the semester with a probability of 0.70 is 274.32**

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