The standard normal distribution and Φ | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Φ(z) is the area to the left of z

The standard normal variable has mean 0 and variance 1: $Z\sim N(0,1)$ . Define

\Phi(z)=P(Z\le z)

Because $Z$ is continuous, $P(Z\le z)=P(Z<z)$ . Move the boundary below and watch the left area change.

upper boundary b = 1.00

P(Z < 1.00) = Φ(1.00)·area = 0.8413

Sketch the bell, 0 and the boundary.
Shade the probability named by the question.
Keep $\Phi(z)$ for a left tail, use $1-\Phi(z)$ for a right tail, or subtract two left areas for an interval.

Read positive, negative and three-decimal z-values

MF19 lists non-negative $z$ -values. For a negative value, use symmetry:

\Phi(-z)=1-\Phi(z)

Worked example

Use the ADD columns for z = 1.234

Read row 1.2 and column 3 first: $\Phi(1.23)=0.8907$ . The third decimal is 4, and the row's ADD entry under 4 is 7, meaning add 0.0007.

\Phi(1.234)\approx0.8914

Answer

\Phi(1.234)\approx0.8914

Hence $\Phi(-1.234)\approx1-0.8914=0.1086$ . A calculator can evaluate the area directly, but the sketch and left/right decision still need to be correct.

Common mistake

\Phi^{-1}

is the inverse function that turns an area into a boundary. It is not the reciprocal

1/\Phi

Use the region to choose the calculation

Left area decides the sign of z

left tail0z

left area 0.90 > 0.5 → z = +1.282

z0right tail

left area 0.20 < 0.5 → z = −0.842

Decide the sign before using the inverse-normal command: a left area above 0.5 places the boundary to the right of 0; a left area below 0.5 places it to the left.

Swipe left or right to read every column →

Required result	Calculator command	Inputs
left area	Normal CD	lower −∞, upper x, mean μ, SD σ
right area	Normal CD	lower x, upper ∞, mean μ, SD σ
area between	Normal CD	lower a, upper b, mean μ, SD σ
boundary z from left area p	Inverse Normal	area p, mean 0, SD 1

Calculator menus use names such as Normal CD, Normal CDF or normalcdf. They mean the same cumulative-area command. Use the largest negative and positive bounds offered by your calculator for $-\infty$ and $\infty$ . If it has no infinity key, typical entries are $-1\times10^{99}$ and $1\times10^{99}$ ; menu labels differ by model.

Worked example

One boundary, two tails

MF19 gives $\Phi(1.23)=0.8907$ .

P(Z<1.23)=0.8907

P(Z>1.23)=1-0.8907

Answer

P(Z>1.23)=0.1093

Worked example

Find a boundary from an area

If $P(Z<a)=0.90$ , the boundary is right of 0 because the left area is greater than 0.5.

a=\Phi^{-1}(0.90)=1.282

Answer

a=1.282

Normal CD is usually the fastest area calculation. Inverse Normal is usually the fastest way to recover a boundary. Write the region and the signed boundary logic before using either command.

Examiner note

MF19 lists only selected common critical areas. Use inverse Normal for any other area. Keep guard digits in the calculator and round only the final answer. A probability must lie between 0 and 1; a right tail beyond a positive boundary must be below 0.5.

Your turnIndependent check [5 marks]

Use the MF19 ADD columns to find $P(-0.842<Z<1.374)$ . Also find $a$ if $P(Z<a)=0.73$ .

Show worked answer

MF19 gives $\Phi(1.37)=0.9147$ ; ADD 4 contributes 0.0006, so $\Phi(1.374)=0.9153$ . It also gives $\Phi(0.84)=0.7995$ ; ADD 2 contributes 0.0005, so symmetry gives $\Phi(-0.842)=0.2000$ .

P=0.9153-0.2000

Answer

P=0.715

a=0.613

Φ(z) is the area to the left of z

The standard normal variable has mean 0 and variance 1: $Z\sim N(0,1)$ . Define

\Phi(z)=P(Z\le z)

Because $Z$ is continuous, $P(Z\le z)=P(Z<z)$ . Move the boundary below and watch the left area change.

upper boundary b = 1.00

P(Z < 1.00) = Φ(1.00)·area = 0.8413

Sketch the bell, 0 and the boundary.
Shade the probability named by the question.
Keep $\Phi(z)$ for a left tail, use $1-\Phi(z)$ for a right tail, or subtract two left areas for an interval.

Read positive, negative and three-decimal z-values

MF19 lists non-negative $z$ -values. For a negative value, use symmetry:

\Phi(-z)=1-\Phi(z)

Worked example

Use the ADD columns for z = 1.234

Read row 1.2 and column 3 first: $\Phi(1.23)=0.8907$ . The third decimal is 4, and the row's ADD entry under 4 is 7, meaning add 0.0007.

\Phi(1.234)\approx0.8914

Answer

\Phi(1.234)\approx0.8914

Hence $\Phi(-1.234)\approx1-0.8914=0.1086$ . A calculator can evaluate the area directly, but the sketch and left/right decision still need to be correct.

Common mistake

\Phi^{-1}

is the inverse function that turns an area into a boundary. It is not the reciprocal

1/\Phi

Use the region to choose the calculation

Left area decides the sign of z

left tail0z

left area 0.90 > 0.5 → z = +1.282

z0right tail

left area 0.20 < 0.5 → z = −0.842

Decide the sign before using the inverse-normal command: a left area above 0.5 places the boundary to the right of 0; a left area below 0.5 places it to the left.

Swipe left or right to read every column →

Required result	Calculator command	Inputs
left area	Normal CD	lower −∞, upper x, mean μ, SD σ
right area	Normal CD	lower x, upper ∞, mean μ, SD σ
area between	Normal CD	lower a, upper b, mean μ, SD σ
boundary z from left area p	Inverse Normal	area p, mean 0, SD 1

Worked example

One boundary, two tails

MF19 gives $\Phi(1.23)=0.8907$ .

P(Z<1.23)=0.8907

P(Z>1.23)=1-0.8907

Answer

P(Z>1.23)=0.1093

Worked example

Find a boundary from an area

If $P(Z<a)=0.90$ , the boundary is right of 0 because the left area is greater than 0.5.

a=\Phi^{-1}(0.90)=1.282

Answer

a=1.282

Normal CD is usually the fastest area calculation. Inverse Normal is usually the fastest way to recover a boundary. Write the region and the signed boundary logic before using either command.

Examiner note

Your turnIndependent check [5 marks]

Use the MF19 ADD columns to find $P(-0.842<Z<1.374)$ . Also find $a$ if $P(Z<a)=0.73$ .

Show worked answer

MF19 gives $\Phi(1.37)=0.9147$ ; ADD 4 contributes 0.0006, so $\Phi(1.374)=0.9153$ . It also gives $\Phi(0.84)=0.7995$ ; ADD 2 contributes 0.0005, so symmetry gives $\Phi(-0.842)=0.2000$ .

P=0.9153-0.2000

Answer

P=0.715

a=0.613