Finding probability given normal distribution

Question

Good day,

Can someone look at the exercise and confirm if my solution is right or correct me if it's wrong?

Answer 1

As Reza Rafie Borujeny says, you seem to be assuming independence.

If so, another approach is to say that $2(X_1+X_2+X_3+X_4)$ is normally distributed with mean $2\times (4\times 3)=24$ and variance $2^2 \times (4 \times 9) = 144 =12^2$.

The Z value is then $\dfrac{48-24}{12}=2$, not you found, and the probability should be about $0.97725$.

The error in your calculations comes from believing the variance of the average of the four $X_i$s is $9$. It is in fact $\frac94 = 2.25$ making the standard deviation $\frac32 = 1.5$. You could read about the standard error of the mean to learn more.

The following R code simulates this:

set.seed(2016)
cases <- 10000000
X1 <- rnorm(cases, mean=3, sd=sqrt(9))
X2 <- rnorm(cases, mean=3, sd=sqrt(9))
X3 <- rnorm(cases, mean=3, sd=sqrt(9))
X4 <- rnorm(cases, mean=3, sd=sqrt(9))
dat <- 2*(X1+X2+X3+X4)

to give a result for the probability of about

> mean(dat < 48)
[1] 0.977292

and for the standard deviation of the average of about

> sd((X1 + X2 + X3 + X4) / 4)
[1] 1.499593

Answer 2

What you wrote is true under the assumption that $X_i$'s are independent random variables. I got the same final answer.