IBDP Math AA Topic: SL 4.2 :Production and understanding of box and whisker diagrams HL Paper 1

Answer:

METHOD 1
\(Q_1\) = 31.86 OR \(Q_3\) = 32.14 recognition that the area under the normal curve below \(Q_1\) or above \(Q_3\) is 0.25 OR the area between \(Q_1\) and \(Q_3\) is 0.5 (seen anywhere including on a diagram)

EITHER
equating an appropriate correct normal CDF function to its correct probability (0.25 or 0.5 or 0.75)

OR
z = −0.674489… OR z = 0.674489… (seen anywhere)
-0.674489… = \(\frac{31.86 – 32}{\sigma }\) OR 0.674489… = \(\frac{32.14 – 32}{\sigma }\)

THEN
0.207564…
\(\sigma = 0.208\) (mm)

METHOD 2
recognition that the area under the normal curve below \(Q_1\) or above \(Q_3\) is 0.25 OR the area between \(Q_1\) and \(Q_3\) is 0.5 (seen anywhere including on a diagram)
z = −0.674489… OR z = 0.674489…
\((Q_1=) 32 – 0.674489… \sigma \) OR \((Q_3 =) 32 + 0.674489… \sigma \)
\((Q_3 – Q_1 =) 2 \times 0.674489… \sigma \)
2 \(\times \) 0.674489… \(\sigma \) = 0.28
0.207564…
\(\sigma = 0.208\) (mm)

The interquartile range (IQR) is a measure of statistical dispersion, which is the difference between the third quartile (Q3) and the first quartile (Q1) of a dataset. For a normal distribution, the IQR is related to the standard deviation \((\sigma)\) by a factor of approximately 1.35.

\(\text{IQR} = Q_3 – Q_1 = 1.349\sigma\)

Where:

• IQR is the interquartile range
• Q3 is the 75th percentile
• Q1 is the 25th percentile
• σ is the standard deviation

The interquartile range of the diameters is 0.28.

We can substitute the known IQR value into the equation: \(0.28 = 1.349\sigma \)

\(\sigma = \frac{0.28}{1.349} \)

\(\sigma \approx 0.208\)