How do you find the probability that the sample mean is between 52 and 56?

Question

I was attempting to help a friend with a question and I am not sure if I am overthinking it, or simply missing an assumption I can make. It goes like this:

The average life of a battery is 50 hours with a standard deviation of 4 hours. What is the probability that the mean of 16 samples is between 52 and 56?

I thought I had this, but I am getting caught up on the fact that it is a small sample size and it is not stated that it is normally distributed.

Question: What method do you use to solve this?

Other people suggested:

$$Z = \frac{\bar{X}-\mu}{\sigma}$$

My problem here is we are assuming it is normal and we are not utilizing the given info about the sample size of 16.

I suggested:

$$Z = \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}$$ My problem here is we are again assuming it is normal and I am getting $2<Z<6$, which gives about $2\%$.

Any clarifications will be greatlyappreciated!

Answer 1

You should not be getting a $97\%$ probability for the mean as (assuming a symmetric distribution) there is a $50\%$ probability the sample mean is below the population mean of $50$

If $Z$ has a standard normal distribution then $P(2 \lt Z \lt 6) \approx 0.999999999 - 0.977249868 \approx 0.02275$, so just over $2\%$

As a simulation in R

set.seed(99)
cases <- 10^5
n <- 16
matdat <- matrix(rnorm(cases*n, mean=50, sd=4), ncol=n)
rm <- rowMeans(matdat)
mean(rm > 52 & rm < 56)

gives 0.02263 - close to correct given simulation error