Answer: 0.8313
Step-by-step explanation:
As per given we have,
[tex]\mu=15.5[/tex] [tex]\sigma= 3.6[/tex]
Also, the distribution of the number of suitcases that get lost each week on a certain route is approximately normal.
Since , [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
z-score corresponds to x= 10:[tex]z=\dfrac{10-15.5}{3.6}\approx-1.53[/tex]
z-score corresponds to x= 20:[tex]z=\dfrac{20-15.5}{3.6}=1.25[/tex]
P-value = [tex]P(10<x<20)=P(-1.53<z<1.25)[/tex]
[tex]=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[/tex]
Hence, the probability that during a given week the airline will lose between 10 and 20 suitcases = 0.8313
