Question 5
A random variable $X$ has mean $\mu$ and standard deviation $\sigma$. The distribution of the sample mean for random samples of size 25 is approximately $\bar{X}\sim N(80,\,4)$.
The value of $\sigma$ is
Not quite. This is the trap. $2=\sqrt{4}$ is the standard deviation of $\bar{X}$, not of $X$. You skipped the step $\dfrac{\sigma}{\sqrt{25}}=2$.
Not quite. You multiplied the variance by $\sqrt{n}$ instead of multiplying the standard deviation by $\sqrt{n}$. That gives $\sigma^{2}=4\times\sqrt{25}=20$, so $\sigma=\sqrt{20}=2\sqrt{5}$.
That’s it. Step through the hints below to see why.
Not quite. You read 4 as the standard deviation of $\bar{X}$, confusing variance with standard deviation. That gives $\sigma=4\times\sqrt{25}=20$.
Read the notation: in $N(\mu,\sigma^{2})$ the second parameter is the variance.
$$\bar{X}\sim N(80,\,4):\quad\text{mean }80,\ \text{variance }4$$
Misreading the 4 as a standard deviation is the main trap in this question.
Recall the distribution of the sample mean for samples of size $n$. Its variance is $\dfrac{\sigma^{2}}{n}$, so its standard deviation is $\dfrac{\sigma}{\sqrt{n}}$.
$$\bar{X}\sim N\!\left(\mu,\frac{\sigma^{2}}{n}\right)$$
This links the spread of $\bar{X}$, which you are given, to the spread of $X$, which you want.
Substitute $n=25$, set the variances equal and solve. Take the positive root, because a standard deviation is positive.
$$\begin{aligned}\frac{\sigma^{2}}{25}&=4\\\sigma^{2}&=4\times25=100\\\sigma&=\sqrt{100}=10\end{aligned}$$
The 4 in $N(80,\,4)$ is the variance of $\bar{X}$, so $\dfrac{\sigma^{2}}{25}=4$. That gives $\sigma^{2}=100$ and $\sigma=10$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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