Answer:
P = 0.0215 = 2.15%
Explanation:
First we need to convert the values of 900 and 975 to standard scores using the equation:
![z = (x - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/high-school/mgftzwxfvlc9iztdu5dhgn7rwdyn3lr15j.png)
Where z is the standard value, x is the original value,
is the mean and
is the standard deviation. So we have that:
standard value of 900:
![z = (900 - 750)/(75) = 2](https://img.qammunity.org/2021/formulas/mathematics/high-school/yajll8panh23jkruea0ir40x3a4yrtwffg.png)
standard value of 975:
![z = (975 - 750)/(75) = 3](https://img.qammunity.org/2021/formulas/mathematics/high-school/8ebls0yhvj4gz5wylv1y3b0d72zdr51vji.png)
Now, we just need to look at the standard distribution table (z-table) for the values of z = 2 and z = 3:
z = 2 -> p_2 = 0.9772
z = 3 -> p_3 = 0.9987
We want the interval between 900 and 975 hours, so we need the interval between z = 2 and z = 3, so we just need to subtract their p-values:
P = p_3 - p_2 = 0.9987 - 0.9772 = 0.0215
So the probability is 0.0215 = 2.15%