Answer:
P = 0.0215 = 2.15%
Step-by-step explanation:
First we need to convert the values of 900 and 975 to standard scores using the equation:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
Where z is the standard value, x is the original value, [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviation. So we have that:
standard value of 900: [tex]z = \frac{900 - 750}{75} = 2[/tex]
standard value of 975: [tex]z = \frac{975 - 750}{75} = 3[/tex]
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%
