Question

he distances in kilometres travelled to work by employees of a city council may be modelled by a normal distribution with mean 75 and standard deviation of 2.5. Find the probability distance travelled to work by a random selected employee of the city council is a) less than 11km b)between 5.5km and 10.5km c)Find d such that 10% of the council’s employees travel less than d kilometers to work.

Answer 1

let, X= Distances in kilometres traveled to work by employees of a city council

$X\backsim N(75,2.5)$

using ,

$z=\frac{x-\bar{x}}{\sigma}=\frac{x-75}{2.5}$ this can be convert to standard normal distribution.

a)

$P(X\le11)=P(Z\le\frac{11-75}{2.5})\\ P(X\le11)=P(Z\le-25.6)\\ P(X\le11)=0$

probability distance traveled to work by a random selected employee of the city council less than 11 km =0

b)

$P(5.5\le X\le10.5)=P(\frac{5.5-75}{2.5}\le Z\le\frac{10.5-75}{2.5})\\ P(5.5\le X\le10.5)=P(-5.5\le Z\le-10.5)\\ P(5.5\le X\le10.5)=\\ \hspace{4 em} P(0\le Z\le-10.5)-P(0\le Z\le-5.5)\\ P(5.5\le X\le10.5)=0-0=0\\$

probability distance traveled to work by a random selected employee of the city council between 5.5 km and 10.5 km =0

c)objective is find a d such that

$P(X\le d)=0.1$

but in standard normal curve,find z such that,

$P(Z\le z)=0.1\\$

$z=-1.282\\ z=\frac{d-75}{2.5}\\ -1.282=\frac{d-75}{2.5}\\ \bold{d=71.795}$