Suppose a sample of 50 is taken from the population with standard deviation 15 and that the sample mean is 100. Establish an interval estimate for the population mean that is a) 90% certain to include the true population mean.

so I have:

n = 50, σ = 15, x̄ = 100

Sample deviation is calculated as population standard deviation divided by square root of sample size, so σx= 12 / sqrt(5) = 1,6

Since the distribution is unknown, what formula should I use?

  • $\begingroup$ You sample size is sufficiently large $(>30)$ so $\overline{X}$ is approximately normal. $\endgroup$
    – user801306
    Commented Jan 11, 2022 at 5:09

1 Answer 1


It seems that you have to assume that the underlying distribution in the population is normal. In your case, the mean is unknown and the variance in known to be 225. In this case, you have $50$ observations of $\mathsf{Norm}(\mu, 225),$ their mean $\bar X$ has distribution $\mathsf{Norm}(\mu, 4.5).$ Then a 90% confidence interval is approximately $$ \mu \in \bar X \pm 1.66 \sqrt{4.5} \approx 100 \pm 1.66 \times 2.13 = 100 \pm 3.54. $$

NOTE: no population is exactly normal, nor exactly any particular distribution. So, never, in practice, you do get a "certain" $\mathsf{CI}$ of any confidence level (except 100% when you provide $\mathbf{R}$).


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