Question 3
In a certain normal distribution curve, 95% of the area lies between the values 50.32 and 113.68. The mean of this distribution is 82.
Determine the standard deviation.
That’s it. Step through the hints below to see why. In 2023, 62.3% of Queensland students chose (A).
Not quite. 21.12 is $31.68\div1.5$, which puts each end 1.5 standard deviations from the mean. That only holds about 86.6% of the area. For 95%, each end is about 1.96 standard deviations out. In 2023, 9.8% of Queensland students chose (B).
Not quite. 31.68 is the distance from the mean to each end. Using it as the standard deviation puts each end one standard deviation away, and only about 68% of the area lies within one standard deviation of the mean. In 2023, 22.9% of Queensland students chose (C).
Not quite. 63.36 is the whole width of the interval, $113.68-50.32$. A standard deviation that big puts each end only half a standard deviation from the mean, which holds about 38% of the area. In 2023, 3.6% of Queensland students chose (D).
Now 95.0% of the area lies between 50.32 and 113.68. Each value is about 1.96 standard deviations from the mean, so $\sigma\approx16.16$.
With $\sigma=21.12$ each value is exactly 1.5 standard deviations from the mean. That holds 86.6% of the area, not 95%.
With $\sigma=31.68$ each value is exactly one standard deviation from the mean. That holds only 68.3% of the area.
With $\sigma=63.36$ each value is only half a standard deviation from the mean. That holds just 38.3% of the area.
This is $31.68\div2$, from the 95% part of the 68–95–99.7% rule. It holds 95.4%, a little more than 95%, and it isn’t an option. The rule rounds 1.96 up to 2.
Very close. The shaded area is within about 0.1% of 95%. Use Check (A) to land on the option value.
More than 95% of the area is shaded, so $\sigma$ is too small. A bigger $\sigma$ spreads the curve out and moves more of the area into the tails.
Less than 95% of the area is shaded, so $\sigma$ is too big. A smaller $\sigma$ pulls more of the area in between the two values.
The mean sits exactly halfway between the two values.
$82-50.32=31.68$$113.68-82=31.68$
So the middle 95% leaves 2.5% in each tail.
The upper value has 97.5% of the area below it. Use the inverse normal with an area of 0.975 to the left.
$\displaystyle P(Z\lt z)=0.975$$\displaystyle \Rightarrow\ z\approx1.96$
So 113.68 is about 1.96 standard deviations above the mean.
$$\frac{113.68-82}{\sigma}=1.96$$
$$\sigma=\frac{31.68}{1.96}\approx16.16$$
The 68–95–99.7% rule rounds 1.96 up to 2, which gives 15.84. That rule is only an approximation: 2 standard deviations either side holds about 95.4%. For exactly 95%, the inverse normal gives 1.96.
$\displaystyle P(Z\lt z)=0.975$$\displaystyle \Rightarrow\ z\approx1.96$
$$1.96=\frac{113.68-82}{\sigma}$$
$$\sigma=\frac{31.68}{1.96}\approx16.16$$
The interval is symmetric about the mean, so each end is 1.96 standard deviations from 82.
How Queensland students answered in 2023
From the QCAA Mathematical Methods subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it