IB Mathematics SL 4.9 normal distribution AA SL Paper 2- Exam Style Questions- New Syllabus

(a)

Using the standard normal distribution table (or a calculator):

\(P(Z<1.7)=0.955434\ldots\)

Hence, to three decimal places,

\(\boxed{P(Z<1.7)=0.955}\)

The cumulative probability gives the area under the normal curve to the left of \(z=1.7\).

✅ Answer: \(0.955\)

(b)

Draw a vertical line at \(z=a\), where \(a\) lies to the left of the mean (\(0\)). Shade the region between \(z=a\) and \(z=1.7\), and label the shaded area as \(0.7865\).

This shaded region represents the probability \(P(a<Z<1.7)\).

✅ Required sketch: Shade the area between \(a\) and \(1.7\).

(c)

Since

\(P(a<Z<1.7)=P(Z<1.7)-P(Z<a)\),

substitute the known values:

\(0.7865=0.955434- P(Z<a)\)

\(P(Z<a)=0.955434-0.7865=0.168934\)

Using the inverse normal distribution (or standard normal tables):

\(a\approx -0.958384\)

Hence,

\(\boxed{a\approx -0.958}\)

The value of \(a\) is negative because the left-tail probability \(0.168934\) is less than \(0.5\), so the corresponding \(z\)-score lies to the left of the mean.

✅ Answer: \(a=-0.958\) (approximately)