SL 4.9 Solve and interpret normal-distribution problems | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Statistics and Probability

IB Mathematics: Analysis and Approaches SL — Study topic SL 4.9

A normal distribution is a model for measurements that cluster around a central value, with fewer observations further from the centre. Possible examples include lengths, times, or scores, when a normal model is appropriate. This lesson connects a context to a symbolic model, a sketch of a bell-shaped curve, and numerical results. Before calculating, decide what event or cutoff the question describes. A sketch can make that decision easier and helps you check the calculator output.

What you will learn

Identify the mean and standard deviation in a normal model and explain their contextual meanings.
Interpret probabilities as areas under a normal curve.
Use standardisation and graphing technology to find probabilities and cutoffs.
State numerical results with suitable accuracy and interpret them in context.

1. Prior knowledge and the normal model

A probability is a number between zero and one that describes how likely an event is. For a continuous measurement, probability is represented by area under a curve. The total area under a normal curve is one, and the curve is symmetric about its mean.

The notation

X\sim N(\mu,\sigma^2)

says that the variable

X

is modelled by a normal distribution with mean

\mu

and standard deviation

\sigma

. The mean gives the centre of the model. The standard deviation describes its spread and has the same units as the measurements. Some calculators ask for standard deviation, while the model notation displays its square, the variance. Check the calculator labels carefully.

A standard score, or

z

-score, tells how many standard deviations a value lies from the mean. A positive score is above the mean; a negative score is below it. Standardising lets us describe values from different normal models using a standard normal distribution with mean zero and standard deviation one.

For a continuous model, the probability of one exact value is zero. Therefore, including or excluding an endpoint does not change an interval probability. In a sketch, mark the mean and relevant boundaries, then shade the requested region.

z=\frac{x-\mu}{\sigma}

The normal curve is symmetric about $\mu$ and has total area $1$ .
The standard deviation $\sigma$ describes spread; it is not the variance $\sigma^2$ .
A probability is an area under the curve, not the curve's height.

2. Probabilities as areas

The words in a question specify a region. “Less than” means shade to the left of the boundary. “Greater than” means shade to the right. “Between” means shade the area between two boundaries. Sketching this region before using a calculator helps prevent choosing the wrong tail.

Let

\Phi(z)

denote the area to the left of

z

under the standard normal curve. The probability between two standardised boundaries is the area to the left of the upper boundary minus the area to the left of the lower boundary. For an original measurement model, you can instead use a cumulative normal calculator function with the original-unit bounds, mean, and standard deviation.

Graphing technology can display the curve and shade an interval or tail, then calculate its area. The display is a useful check, but the mathematical reasoning still matters: identify the event and select the correct bounds. For a left-tail probability, the lower bound is negative infinity; for a right-tail probability, the upper bound is positive infinity. If a calculator does not accept infinity, use an appropriately large negative or positive bound.

A probability must be between zero and one. A left-tail probability and its complementary right-tail probability add to one. Retain calculator precision during working, and round the final result to the requested accuracy.

P(a<X<b)=\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)

Translate the wording into a shaded region before calculating.
Use original units with the model parameters, or standardise both bounds consistently.
Check the result's size against the sketch and the probability range.

3. Finding a value from a probability

Some questions give an area or percentage and ask for the measurement at its boundary. For instance, a cutoff with 90% of values below it has left-tail area

0.90

. Finding a boundary from a cumulative area is an inverse normal calculation.

Use the inverse normal function with the area to the left, mean, and standard deviation. If the function returns a standard score, convert it back to the measurement scale by multiplying by the standard deviation and adding the mean. For a right-tail percentage, first subtract that percentage from one to obtain the area to the left.

Interpret the returned value in the original units. A cutoff is a measurement, not a probability. Keep sufficient calculator precision until the final step, especially if the answer is to be rounded to a whole unit. After rounding, the stated percentage may be approximate rather than exact.

x=\mu+z\sigma

An inverse normal calculation starts with a cumulative area and returns a boundary.
Check whether the question gives an area to the left or to the right.
Use the mean and standard deviation in the same units as the requested value.

4. A reliable solving routine

Start by naming the variable, its model, and the mean and standard deviation with units. Decide whether the question asks for a probability or a measurement cutoff. Draw a small labelled sketch and shade the region or indicate the cumulative area.

For a probability, calculate the area of the shaded region. For an unknown cutoff, use the correct cumulative area in an inverse normal calculation. Then write a contextual answer, including units for a measurement but not for a probability.

Use technology purposefully: a graph can confirm which region is being measured, and a calculator can evaluate the numerical area or inverse value. Check whether the result is on the expected side of the mean and whether its size agrees with the sketch. These checks can reveal a wrong tail, swapped bounds, or entering variance where standard deviation is required.

Model, sketch, calculate, interpret, and check.
Probabilities have no units; measurement cutoffs retain the original units.
Calculator output should support, not replace, an explanation of the event.

Worked example

A right-tail probability

A filling machine produces packets whose mass is modelled by

X\sim N(500,12^2)

grams. Find the probability that a packet has mass greater than

518

grams. Give the answer to three significant figures.

Identify the event
The mean is $500$ grams and the standard deviation is $12$ grams. “Greater than” asks for the area to the right of $518$ grams, so the probability should be less than one half.
$P(X>518)$
Standardise the boundary
The boundary is $18$ grams above the mean. Dividing by the standard deviation expresses that distance in standard-deviation units.
$z=\frac{518-500}{12}=1.5$
Calculate and interpret
Use the standard normal right-tail area, or subtract the cumulative area to the left from one. The result is approximately $0.0668$ , consistent with a relatively high boundary.
$P(Z>1.5)\approx 0.0668$

Answer: The probability is approximately

0.0668

, or 6.68%.

Check: The boundary is above the mean, so the right-tail probability is below one half.

Worked example

A probability between two values

The time to complete a task is modelled by

T\sim N(42,5^2)

minutes. Find the probability that a randomly selected completion time is between

37

and

49

minutes. Give the answer to three significant figures.

Represent the interval
The event is the area between $37$ and $49$ minutes. Both bounds use the same units as the mean and standard deviation.
$P(37<T<49)$
Standardise both bounds
The lower bound is one standard deviation below the mean, and the upper bound is $1.4$ standard deviations above it.
$z_1=\frac{37-42}{5}=-1,\qquad z_2=\frac{49-42}{5}=1.4$
Find the area
Subtract the cumulative area to the left of the lower score from the cumulative area to the left of the upper score. Using standard normal values gives $0.919243-0.158655=0.760588$ .
$P(-1<Z<1.4)\approx 0.760588$

Answer: The probability is approximately

0.761

.

Check: The interval contains the mean and extends farther above it than below it, so an area greater than one half is reasonable. The numerical value is the difference of the two cumulative areas.

Worked example

Finding a cutoff from a percentile

Scores on a placement test are modelled by

S\sim N(68,9^2)

points. Find the score exceeded by only the highest 10% of test takers. Give the cutoff to the nearest whole point.

Convert to a left-tail area
If only the highest 10% exceed the cutoff, then 90% are below it. Use a left-tail area of $0.90$ in the inverse normal function.
$P(S<x)=0.90$
Find the standard score
The inverse standard normal value for a left-tail area of $0.90$ is approximately $1.2816$ . It is positive, as the cutoff is above the mean.
$z\approx 1.2816$
Convert to score units
Multiply the standard score by the standard deviation and add the mean. Round only after finding the score in points.
$x=68+(1.2816)(9)\approx79.534$

Answer: The cutoff is

80

points to the nearest whole point.

Check: The cutoff is above the mean, and the area to its right is

0.10

, as required.

Common mistakes and how to avoid them

Entering the variance when the calculator asks for standard deviation.

Correction: For

N(\mu,\sigma^2)

, enter

\sigma

in a field labelled standard deviation.

Using a left-tail area directly for a “greater than” question.

Correction: Sketch the right tail. If the calculator returns the area to the left, subtract that value from one.

Treating a highest-percent statement as the left-tail percentage.

Correction: Convert the wording first: “highest 10%” means a right-tail area of

0.10

and a left-tail area of

0.90

.

Giving measurement units to a probability, or omitting units from a cutoff.

Correction: Probabilities are unitless. State a boundary in the original measurement units.

Lesson summary

A normal model is described by its mean and standard deviation; its curve is symmetric and has total area one.
Probabilities are areas. Sketch the requested event to identify the correct tail or interval.
Standardise with $z=(x-\mu)/\sigma$ when useful, or use a cumulative calculator function with original units and model parameters.
For an unknown boundary, use inverse normal with the correct left-tail area, then interpret the value in context.
Check the input parameters, direction, units, and approximate size of the result.

Check your understanding

Question 1

If

X\sim N(30,4^2)

, what is the standardised value of

x=38

?

$z=2$
$z=-2$
$z=8$
$z=0.5$

Show answer and explanation

z=2

The value is

8

above the mean, which is

8/4=2

standard deviations above it.

Question 2

A calculator returns

P(X<a)=0.73

. What is

P(X>a)

?

$0.27$
$0.73$
$1.73$
$0.50$

Show answer and explanation

0.27

The complementary areas add to one, so the right-tail probability is

1-0.73=0.27

.

Question 3

A cutoff has

0.25

of the distribution below it. Relative to the mean, where must it lie?

Below the mean
Above the mean
Exactly at the mean
Its position cannot be determined

Show answer and explanation

Below the mean

By symmetry, the mean has one half of the distribution below it. A left-tail area of

0.25

therefore has its cutoff below the mean.

Key terms

Normal distribution: A symmetric, bell-shaped probability model described by a mean and standard deviation.
Mean: The central value of a normal model, written as $\mu$ .
Standard deviation: A measure of spread in the same units as the data, written as $\sigma$ .
Standard score: The number of standard deviations a value is from the mean, calculated by $z=(x-\mu)/\sigma$ .
Percentile: A value with a stated proportion of the distribution at or below it.

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SL 4.9 · Solve and interpret normal-distribution problems

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 4.9

What you will learn

1. Prior knowledge and the normal model

2. Probabilities as areas

3. Finding a value from a probability

4. A reliable solving routine

A right-tail probability

A probability between two values

Finding a cutoff from a percentile

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review