Question 3 2 pts (CO 3) The monthly utility bills in a city are normally distributed with a mean of $\$ 121$ and a standard deviation of $\$ 41$. Find the probability that a randomly selected utility bill is between $\$ 110$ and $\$ 130$.

Step 1 ：Calculate the z-score for $110 using the formula: \(z_1 = \frac{110 - 121}{41} = -0.268\)

Step 2 ：Calculate the z-score for $130 using the formula: \(z_2 = \frac{130 - 121}{41} = 0.220\)

Step 3 ：Find the probability corresponding to \(z_1\) using a standard normal distribution table or calculator: approximately 0.3950

Step 4 ：Find the probability corresponding to \(z_2\) using a standard normal distribution table or calculator: approximately 0.5871

Step 5 ：Subtract the probability corresponding to \(z_1\) from the probability corresponding to \(z_2\) to find the probability that a randomly selected utility bill is between $110 and $130: \(0.5871 - 0.3950 = 0.1921\)

Step 6 ：The probability that a randomly selected utility bill is between $110 and $130 is approximately 0.1921 or 19.21%