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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)

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

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

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  • $\begingroup$ 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. $\endgroup$ – Alex Ionovich Page Nov 2 '17 at 22:39
  • $\begingroup$ In the first part, you conclude that $Var(\bar{X}) = \frac{\sigma^2}{n}$, so just take square root. $\endgroup$ – Siong Thye Goh Nov 2 '17 at 22:42

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