Question

Assume that 5% of a certain model of tires wear out before 25,000 miles, and another 5% of the tires of that model exceed 35,000 before wearing out. What percentage of tires are still working at 24,000 miles if wear out time is normally distributed?

Answer 1

let mean and standard deviation are a and b

for tail probability of 5%, critical z =-/+ 1.645

therefore a-1.645b=25000 ..........(!)

a+1.645b=35000 ...........(2)

adding (1) and (2)

a =mean =(35000+25000)/2 =3000

and standard deviation=(35000-30000)/1.645 =3039.51

therefore from normal distribution :

percentage of tires are still working at 24,000 miles :

probability =P(X>24000)=P(Z>(24000-30000)/3039.514)=P(Z>-1.97)=1-P(Z<-1.97)=1-0.0244=0.9756~ 97.56%