mean and standard derivation about battery The batteries produced in a factory are tested before packing: 1.5% of batteries are found to be faulty, and are scrapped. Whether or not a battery is faulty is independent of each other. Experience suggested that a ‘good’ battery could last for 36 to 45 hours when used, and that all times within this range are equally likely.

D: The total time that a box of 100 good batteries lasts when used.

(c) State the distribution, including the values of any parameters, that may be used to approximate the distribution of D, the total time that a box of 100 good batteries lasts. Also, Justify the use of this distribution briefly. 

(d) Use the distribution that you have suggested in part c to calculate an approximate value for the probability that a box of 100 good batteries will last for less than 4000 hours. 

i can't find out the mean and standard derivation of it, moreover, can i use normal distribution to approximate the probability, can someone help me?
Background. I suppose the $n = 100$ batteries are used sequentially, so that
the total time the box lasts is $T = X_1 + X_2 + \cdots +  X_{100},$
where $X_i \sim \mathsf{Unif}(35.45).$ 

By the Central Limit Theorem, the sum of such a large number of
independent uniformly distributed random variables is very
nearly normally distributed. (In fact, only $n=10$ would be large enough to get a nearly normal total. You may want to look at Wikipedia on the 'Irwin-Hall Dist'n'.)

In order to work the problem, you need to find $E(X_i) = 40.5.$ and also $Var(X_i).$ Then use these results to find $E(T)$ and $Var(T).$
Then the final answers use a normal distribution, with that mean and variance.

With that start, I will leave the details to you. This seems to be
a homework problem. Soon, there may be a similar problem on an
exam, and I want you to know how to do that. I will simulate the
answers, using R statistical software, so you will have something to check against when you've
finished working this problem. 

Notes: (1) A simulation based on a million boxes
will give pretty good approximate answers. (2) If you are using printed tables of the standard normal CDF, that may involve some rounding, so your probability answer may
differ slightly from mine.

Simulation. Total times for $m = 10^6$ boxes, each with $n = 100$ batteries are simulated using R statistical software. 

set.seed(2020)
t = replicate(10^6, sum(runif(100,36,45)))
mean(t);  sd(t)
[1] 4050.054     # aprx E(T) = 4050
[1] 25.98528     # aprx SD(T) = 25.98
mean(t &lt; 4000)
[1] 0.026872     # aprx P(T &lt; 4000) = 0.027
pnorm(4000, 4050, 25.98)
[1] 0.02714238   # P(T &lt; 4000)


Here is a histogram of the lifetimes $T$ of a million boxes along
with the density curve of the approximate normal distribution of $T.$ The area under the curve to the left of the vertical broken
line represents $P(T &lt; 4000).$

hist(t, prob=T, col="skyblue2", 
   main="Simulated Dist'n of Box Lifetime")
 curve(dnorm(x, 4050, 25.98), add=T, lwd=2)
 abline(v = 4000, col="red", lwd=2, lty="dotted")
(c) We will use the continuous uniform distribution to approximate the distribution of D. The total time that a 100 good batteries can last is in the interval of [36 * 100, 45 * 100] = [3600, 4500]. Because every good battery is equally likely to last from 36 to 45, the total time of 100 batteries is also equally likely to be in the range of 3600 to 4500. That is why we use the continuous uniform distribution.

X - total lasting time of a 100 good batteries

probability density function: f(x) = 1/(b - a) , a &lt;= x &lt;= b

a = 3600
b = 4500

mean = E(X) = (a + b)/2

variance = V(X) = (b - a)^2 / 12

(d) The cumulative distribution function is: $F(x) = \int_{a}^{x} 1 /(b - a)du = x/(b - a) - a/(b - a)$

now we only substitute for our values

F(4000) = 4000/(4500 - 3600) - 3600/(4500 - 3600) = 4000/900 - 3600/900 = 4.(4) - 4 = 0.(4)