Answer to Question #147352 in Statistics and Probability for Rehan Tahir

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
iii)between 10 and 12
iv)between 8 and 10
v)between 12 and 14

Answer 1

$\mu=10,\ \sigma=2$

a) The values between 8 and 12 are $1\sigma$ away from the mean thus due to the 68-95-99 rule the percentage of such motors is 68%

2) $P(6<X<8)=P(\frac{6-10}{2}<Z<\frac{8-10}{2})=P(-2<Z<-1)=$

$=P(Z<-1)-P(Z<-2)=0.1587-0.0228=0.1359=13.59\%$

3) $P(10<X<12)=P(\frac{10-10}{2}<Z<\frac{12-10}{2})=P(0<Z<1)=$

$=P(Z<1)-P(Z<0)=0.8413-0.5=0.3413=34.13\%$

Or values between 10 and 12 are $1\sigma$ away to the right from the mean thus due to the 68-95-99 rule the percentage of such motors is 68%/2 = 34%

4) Values between 8 and 10 are $1\sigma$ away to the left from the mean thus due to the 68-95-99 rule the percentage of such motors is 68%/2 = 34%

5) $P(12<X<14)=P(\frac{12-10}{2}<Z<\frac{14-10}{2})=P(1<Z<2)=$

$=P(Z<2)-P(Z<1)=0.9772-0.8413=0.1359=13.59\%$

Answer:

i) 68%

ii) 13.59%

iii) 34%

iv) 34%

v) 13.59%