Answer to Question #147711 in Statistics and Probability for Aina

Question

The average life of certain type of small motors is 10 years with standard deviation of 2 years. Suppose that the life of motors is normally distributed. Then what percentage of motors will

i) Between 8 and 12

ii) Between 6 and 8

between 10 and 12

between 8 and 10

between 12 and 14

Answer 1

1) Between 8 and 12

P(8<x<12)=P(x<12)-P(x<8)

P(x<8)

z=$\frac{x-\mu}{\sigma}$ =(8-10)/2=-1

using table, P(z=-1)=0.15866

P(x<12)

z=(12-10)/2=1

P(z=1)=0.84134

P(8<x<12)=0.84134-0.15866=0.68268

2) between 6 and 8

P(6<x<8)=P(x<8)-P(x<6)

P(x<8)=0.15866

P(x<6)

z=(6-10)/2=-2

P(z=-2)=0.02275

P(6<x<8)=0.15866-0.02275=0.13591

3) between 10 and 12

P(10<x<12)=P(x<12)-P(x<10)

P(x<12)=0.84134

P(x<10)

z=(10-10)/2=0

P(z=0)=0.5

P(10<x<12)=0.84134-0.5=0.34134

4)between 8 and 10

P(8<x<10)=P(x<10)-P(x<8)

P(x<8)=0.15866

P(x<10)=0.5

P(8<x<10)=0.5-0.15866=0.34134

5)between 12 and 14

P(12<x<14)=P(x<14)-P(x<12)

P(x<12)=0.84134

P(x<14)

z=(14-10)/2=2

P(z=2)=0.97725

P(12<x<14)=0.97725-0.84134=0.13591