IB Mathematics SL 4.12 Standardization of normal variables AA SL Paper 1- Exam Style Questions- New Syllabus

(a)
Since \(X\) and \(Y\) have the same standard deviation \(a\), we can compare their standardized Z-scores.
For the probabilities to be equal, the Z-score for \(X > b\) must equal the Z-score for \(Y > 22\).
\(Z_X = \frac{b – \mu_X}{\sigma} = \frac{b – 7}{a}\)
\(Z_Y = \frac{22 – \mu_Y}{\sigma} = \frac{22 – 19}{a}\)
Equating the Z-scores:
\(\frac{b – 7}{a} = \frac{3}{a}\)
\(b – 7 = 3\)
\(b = 10\)

(b)
The interval \(7 – a < X < 7 + a\) corresponds to \(\mu – \sigma < X < \mu + \sigma\).
For a normal distribution, approximately \(68\%\) of the data lies within one standard deviation of the mean.
Probability \(\approx 0.68\)

(c)
Given \(a = 3\), \(Y \sim N(19, 3^2)\).
We want to find \(P(Y < 22)\).
Standardize to find the Z-score:
\(Z = \frac{22 – 19}{3} = \frac{3}{3} = 1\)
\(P(Y < 22) = P(Z < 1)\)
Using the symmetry of the normal curve and the result from (b):
\(P(Z < 1) = 0.5 + \frac{0.68}{2} = 0.5 + 0.34 = 0.84\)