Answer:
A
Explanation:
From the given information;
The required probability needed to carry out is P(32000<X<42000);
Given that:
mean [tex]\mu[/tex] = 40000
standard deviation [tex]\sigma[/tex] = 5000
Using the standard normal distribution;
[tex]P(32000 <X<42000) = ( \dfrac{x - \mu}{\sigma} <Z< \dfrac{x - \mu}{\sigma})[/tex]
[tex]P(32000 <X<42000) = ( \dfrac{32000 - 40000}{5000} <Z< \dfrac{42000 - 40000}{5000})[/tex]
[tex]P(32000 <X<42000) = ( -1.6<Z<0.4)[/tex]
Here, the region of the area lies between -1.60 and 0.40
∴
P(320000 < X < 40000) = P(Z<0.40) - P(Z< -0.40)
From Z tables;
P(320000 < X < 40000) = 0.6554 -0.0548
P(320000 < X < 40000) = 0.6006
P(320000 < X < 40000) = 60.06%