I need help with part C please help. **Study Summary:**A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 142.6 seconds. Assuming drive-through times are normally distributed with a standard deviation of 25 seconds, complete parts (a) through (d) below.**Resources:**- [Click here to view the standard normal distribution table (page 1)](#)- [Click here to view the standard normal distribution table (page 2)](#)---**Questions and Answers:****(a)** What is the probability that a randomly selected car will get through the restaurant's drive-through in less than 97 seconds?- The probability that a randomly selected car will get through the restaurant's drive-through in less than 97 seconds is **0.0344**. (Round to four decimal places as needed.)**(b)** What is the probability that a randomly selected car will spend more than 173 seconds in the restaurant's drive-through?- The probability that a randomly selected car will spend more than 173 seconds in the restaurant's drive-through is **0.1119**. (Round to four decimal places as needed.)**(c)** What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through?- The proportion of cars that spend between 2 and 3 minutes in the restaurant's drive-through is **[Blank Box]**. (Round to four decimal places as needed.)

GivenMean(μ)=142.6standard deviation(σ)=25c) Proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through is2min=2*60=120 sec3min=3*60=180 sec=P(120<x<180)=P(120-μσ<x-μσ<180-μσ)=P(120-142.625<z<180-142.625)=P(-0.904<z<1.496)=P(z<1.50)-P(z<-0.90)=0.9332-0.1841 (from z table)=0.7491The Proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through is 0.7491