Question

n airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with μ = 15.5 and σ = 3.6. What is the probability that during a given week the airline will lose between 10 and 20 suitcases?

Answer 1

Answer: 0.8313

Step-by-step explanation:

As per given we have,

$\mu=15.5$  $\sigma= 3.6$

Also, the distribution of the number of suitcases that get lost each week on a certain route is approximately normal.

Since , $z=\dfrac{x-\mu}{\sigma}$

z-score corresponds to x= 10:$z=\dfrac{10-15.5}{3.6}\approx-1.53$

z-score corresponds to x= 20:$z=\dfrac{20-15.5}{3.6}=1.25$

P-value = $P(10<x<20)=P(-1.53<z<1.25)$

$=P(z<1.25)-P(z<-1.53)\\\\=P(z<1.25)-(1-P(z<1.53))\\\\=0.8943502-(1-0.9369916)=0.8313418\approx0.8313$

Hence, the probability that during a given week the airline will lose between 10 and 20 suitcases = 0.8313