**7. Correct. The answer is true**

Let

m_{A} = 15 years, m_{B} = 7 years

s_{A} = 5 years, s_{B} = 2 years

n_{A} = 10, n_{B}
= 10

X_{A} - X_{B} = 8 years

Difference in means
is given by;

m_{A-B} = 15 - 7 = 8 years

Difference in
standard deviation is given by

Standardized
variable for the difference in the means is given by;

Z = [(X_{A}
- X_{B}) - (m_{A} - m_{B})] / s_{A-B}

Z = [10-8] / 1.703 =
1.17

Required probability
= (area under normal curve to the right of Z = 1.17)

= 0.5 - 0.3790 =
0.121

Therefore, the
probability that the Chilean copper mines have a mean lifetime at least 10 years
greater than that of their Australian counterparts is 0.121.