IBDP Maths SL 4.12 Standardization of normal variables AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
Let \(L\) represent the height of a large sunflower.

Since \(25\%\) of the sunflowers have a height greater than \(180\text{ cm}\):

\(P(L>180)=0.25\)

Therefore:

\(P(L\leq180)=0.75\)

The standardized value corresponding to a cumulative probability of \(0.75\) is:

\(A=\Phi^{-1}(0.75)=0.674489\ldots\)

Thus:

\(\dfrac{180-\mu}{\sigma}=0.674489\ldots\)

or equivalently:

\(\mu+0.674489\ldots\sigma=180\)

✅ Answer: \(A=0.674\)

(b)
From part (a):

\(\mu+0.674489\sigma=180\)

The giant sunflowers have mean \(\mu+35\) and standard deviation \(2\sigma\).

Let \(G\) represent the height of a giant sunflower. Since \(98\%\) have a height greater than \(180\text{ cm}\):

\(P(G>180)=0.98\)

Therefore:

\(P(G\leq180)=0.02\)

The standardized value corresponding to a cumulative probability of \(0.02\) is:

\(\Phi^{-1}(0.02)=-2.053748\ldots\)

Hence:

\(\dfrac{180-(\mu+35)}{2\sigma}=-2.053748\ldots\)

This gives:

\(180=\mu+35-4.107497\ldots\sigma\)

Using \(\mu=180-0.674489\ldots\sigma\):

\(180=180-0.674489\ldots\sigma+35-4.107497\ldots\sigma\)

\(4.781987\ldots\sigma=35\)

\(\sigma=7.31913\ldots\)

Now substitute this into \(\mu+0.674489\ldots\sigma=180\).

\(\mu=180-0.674489\ldots(7.31913\ldots)\)

\(\mu=175.063\ldots\)

Therefore, to three significant figures:

✅ Answer: \(\mu=175\text{ cm}\) and \(\sigma=7.32\text{ cm}\)