Normal distribution example

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;

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

$\mu =226 and the \delta =16$

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

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

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

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

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

=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?

$U\sin g Normal Distribution \mu =266 and \delta =16$

$P(x<?)=0.70$

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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