PLEASE RE-SOLVE THIS QUESTION SINCE I PREVIOUSLY RECEIVED WRONG ANSWERS FROM CHEGG

A brief refresher on Rules for Means and Variances. Suppose Di is a random variable, where i is a value between 1 and N. The mean (or expected value) of Di is denoted E[Di] and the variance of Di is denoted Variance[Di]. When D1,D2,,DN are uncorrelated (i.e., for each pair of random variables (Di,Dj,), the correlation is zero), then we have the following Rules for Means and Variances: Variance[D1 + D2 ++ DN] = Variance[D1] + Variance[D2] ++ Variance[DN] (The equation for variance has been simplified to drop out the covariance terms because in our setting, by assumption, the random variables are uncorrelated.)

E[D1 + D2 ++ D N] = E[D1] + E[D2] ++ E[DN].

A family plan for 800 megabytes of mobile data is shared by three people. Each family members monthly data usage in megabytes is normally distributed with the following mean and variance:

Parent #1, Mean: 200, Variance: 3000

Parent #2, Mean: 400, Variance: 5000

Kid, Mean: 156, Variance: 2000

1. What is the mean of the number of megabytes of data the family uses in a month? (I previously received an answer of 306 from Chegg, but it is the WRONG answer, please RE-SOLVE the question)
2. What is the standard deviation of the number of megabytes of data the family uses in a month? (I previously received an answer of 282.84 from Chegg, but it is the WRONG answer, please RE-SOLVE the question)
3. What is the probability that the family will not run out of its monthly plan megabytes (hint: you will need to consult a Normal Table to answer this question)? (I previously received an answer of 95.99% from Chegg, but it is the WRONG answer, please RE-SOLVE the question)
4. Suppose the family can choose the number of megabytes in their plan at the start of the month. How many megabytes should they choose if they want to be 97.5% certain that they would not run out of their megabytes (i.e. they want to make sure that they will not run out of megabytes with probability 0.975)? (I previously received an answer of 557.28 from Chegg, but it is the WRONG answer, please RE-SOLVE the question)