Number of sample units to have a given probability that the sample mean is less than a value

Question

Would this be the correct approach to finding $n$ if I want the sample mean to be less than $5.7$ with a probability of $95\%$? ( Given $X \sim \mathcal{N}(\mu=6,\,\sigma^{2}=0.3^{2})$ )

$Pr(\bar{X}<5.7)=0.95$

$Pr\left(z<\frac{5.7-6}{0.3/\sqrt{n}}\right)=0.95$

$\frac{5.7-6}{0.3/\sqrt{n}}=1.65$

Thank you.

Answer 1

I think you have it backwards. If the population mean is $\mu = 6,$ then it will be difficult for the mean of a large-ish sample to be less than $5.7.$ In your formulation it will require a negative z-value to match the negative quantity $\frac{5.7-6}{0.3/\sqrt{n}}.$

If you want the sample mean of $n$ observations to be more than 5.7 about 95% of the time, then you have $-1.645 = \frac{5.7-6}{0.3/\sqrt{n}}.$ And, solving for $n$ we get $n \approx 1.645^2 = 2.7055,$ so $n =3$ observations should suffice.

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Note: In case you are still having trouble with the intuition of this, here is a simulation in R statistical software with $m = 10^6$ samples of size $n,$ one sample of size $n = 3$ on each row of the matrix. Of the one million sample averages a simulated in this way, almost 96% exceeded $5.7.$ (Being slightly fussier about the simulation result, the expected proportion of averages exceeding $5.7$ is about $0.95807 \pm 0.00040.$)

m = 10^6;  n = 3
x = rnorm(m*n, 6, .3)
MAT = matrix(x, nrow=m)
a = rowMeans(MAT)
mean(a > 5.7)
## 0.95807             # simulated value
2*sd(a > 5.7)/sqrt(m)
## 0.0004008587        # 95% margin of simulation error

1-pnorm(5.7, 6, .3/sqrt(3))
## 0.9583677           # exact probability

In the figure below the histogram shows the simulated sample means, the curve is the density function of $\mathsf{Norm}(\mu=6, \sigma = .3/\sqrt{3}),$ and the area under the density curve to the right of the vertical dotted line is about 0.958.