# Where did I go wrong with this Sampling Question?

Q) Random Samples of three are drawn from a population of beetles whose lengths have a normal distribution with mean $2.4cm$ and standard deviation $0.36cm$. The mean length $\bar {X}$ is calculated for each sample.

i) State the distribution of $\bar {X}$ giving the values of its parameters.

$$X(2.4,0.36^2/3)$$ $$X(2.4,0.0432)$$ $$(\mu = 2.4, \sigma^2 = 0.0432)$$

(This is correct)

ii) Find $P(\bar {X}>2.5)$

$$\implies 1- P(\bar {X} \le 2.5)$$ $$Z = X-\mu/(σ/n)$$ $$Z = \frac{2.5-2.4}{0.36/3}$$ $$Z = 0.833$$ $$\implies 1- \Phi(0.833)$$ $$= 1- 0.7975$$ $$= 0.2025$$

(This answer is wrong, the right answer is 0.3152)

## 1 Answer

In the denominator $\frac{0.36}{3}$ should be $\frac{0.36}{\sqrt3}$

It should be $\frac{\sigma}{\sqrt{n}}$ rather than $\frac{\sigma}{n}$

• Thank you, why is n rooted? I suppose that this is the case w. all sampling/normal distribution questions when finding prob of the mean is larger/smaller than a value. – Alex Ionovich Page Nov 2 '17 at 22:39
• In the first part, you conclude that $Var(\bar{X}) = \frac{\sigma^2}{n}$, so just take square root. – Siong Thye Goh Nov 2 '17 at 22:42