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Topic 4.10 · SL and HL

Normal distribution + bell curve (+inv normal): notes and practice questions

Summary
  • Normal distribution is a continuous probability distribution.
  • Notation: X∼N(μ,σ2)X \sim N(\mu, \sigma^2), where μ\mu is the population mean and σ2\sigma^2 is the population variance.
  • The distribution is symmetrical, bell-shaped, and has a single mode.
  • Changing μ\mu translates the entire graph horizontally.
  • Changing σ2\sigma^2 alters the curve's shape: small σ2\sigma^2 results in a tall, narrow curve; large σ2\sigma^2 results in a short, wide curve.
  • The total area under the normal curve is always 1.
  • 68-95-99.7 Rule:
  • 68% of data lies within one standard deviation of the mean (μ±σ\mu \pm \sigma).
  • 95% of data lies within two standard deviations of the mean (μ±2σ\mu \pm 2\sigma).
  • 99.7% of data lies within three standard deviations of the mean (μ±3σ\mu \pm 3\sigma).
  • For any continuous distribution, the probability of the random variable taking a single exact value is zero: P(X=x)=0P(X = x) = 0.
  • Consequently, P(X≤x)=P(X<x)P(X \le x) = P(X < x) and P(X≥x)=P(X>x)P(X \ge x) = P(X > x).
  • To calculate probabilities for a range (P(a<X<b)P(a < X < b)), use your GDC's Normal Cumulative Distribution (NCD) function with lower bound (aa), upper bound (bb), mean (μ\mu), and standard deviation (σ\sigma).
  • If given variance (σ2\sigma^2), always calculate standard deviation as σ=σ2\sigma = \sqrt{\sigma^2} for GDC input.
  • For P(X>a)P(X > a), use a sufficiently large upper bound (e.g., 109910^{99}) in NCD.
  • For P(X<b)P(X < b), use a sufficiently small lower bound (e.g., −1099-10^{99}) in NCD.
  • Inverse Normal Distribution: Used to find the data value (aa) when the probability (area) is given.
  • For P(X<a)=pP(X < a) = p: Use GDC's Inverse Normal (InvN) function with area (pp), mean, and standard deviation (left tail).
  • For P(X>a)=pP(X > a) = p: If GDC has a 'right tail' option, use it. Otherwise, use the complement rule P(X<a)=1−pP(X < a) = 1 - p and input 1−p1-p as the area for a 'left tail' calculation.
  • Always use NCD for calculating probabilities and InvN for working backwards from probabilities; ignore Normal Probability Density (NPD).
  • Always input the standard deviation (σ\sigma) into the GDC, never the variance (σ2\sigma^2).
  • Sketch a bell curve to visualize the probability area or tail.
  • Check the sense of inverse normal answers (e.g., if P(X<a)<0.5P(X < a) < 0.5, then aa must be less than μ\mu).
  • P(a≤X<b)P(a \le X < b) is evaluated identically to P(a<X<b)P(a < X < b) due to P(X=x)=0P(X=x)=0.
  • HL students also cover linear combinations of normal random variables and using sample mean distributions for Confidence Intervals.

How it is examined

The exclusion of zz-scores is the single most useful fact here, and it is the biggest difference from AA. Everything is done on the GDC with μ\mu and σ\sigma given. Inverse normal questions give the mean and standard deviation and ask for a cut-off, which is the harder direction for students because the calculator needs a tail specified. Sketching the curve with the region shaded is often worth a mark, so this subtopic wants drawing support.

Key ideas
  • The normal distribution and curve.
  • Properties of the normal distribution.
  • Diagrammatic representation.
  • Normal probability calculations.

Linking questions

  • Links to other subjects: normally distributed real-life measurements and descriptive statistics (sciences, psychology, environmental systems and societies).
  • Aim 8: why might the misuse of the normal distribution lead to dangerous inferences and conclusions?
  • International-mindedness: De Moivre's derivation of the normal distribution and Quetelet's use of it to describe l'homme moyen.
  • TOK: to what extent can we trust mathematical models such as the normal distribution? How can we know what to include, and what to exclude, in a model?

Practice questions

42 questions · 25 medium · 17 hard
Showing 20 of 20

Question 1

MediumPaper 1 · calculator5 marks
(a)

The volume of soda in bottles produced by a certain machine is normally distributed with a mean of 330 ml and a standard deviation of 2.5 ml.

(a) Bottles are classified as 'overfilled' if their volume is between 333 ml and 336 ml. Determine the probability that a randomly selected bottle is overfilled.

[2]
(b)

(b) Approximately 75% of bottles are classified as 'standard fill', which means their volume is between kk ml and 333 ml.

Find the value of kk.

[3]

Question 2

HardPaper 1 · calculator7 marks
(a)

A beverage company produces bottles of orange juice. The volume of juice in each bottle, in mL, can be modelled by a normal distribution with a mean of 1005 mL and a standard deviation of 12 mL.

Find the probability that a randomly selected bottle contains less than its labelled volume of 1000 mL.

[2]
(b)

Find the upper quartile of the volumes of the bottles.

[2]
(c)

The bottles are packed into cartons, with 8 bottles in each carton. The volumes of juice in the bottles are independent of each other.

Find the probability that the total volume of juice in a carton exceeds 8050 mL.

[3]

Question 3

MediumPaper 1 · calculator6 marks
(a)

[Maximum mark: 6]

The lifespan of a certain brand of LED light bulbs is approximated by a normal distribution with a mean of 1500 hours and a standard deviation of 120 hours.

A light bulb from this brand is chosen at random.

(a) Calculate the probability that the lifespan of the light bulb is less than 1350 hours.

[2]
(b)

It is known that 25% of the light bulbs have a lifespan greater than kk hours.

(b) Find the value of kk.

[2]
(c)

For a randomly chosen light bulb with lifespan LL hours, P(1500−m<L<1500+m1500 - m < L < 1500 + m) = 0.65.

(c) Find the value of mm.

[2]

Question 4

HardPaper 1 · calculator12 marks
(a)

The lifespan of a certain brand of smartphone battery is normally distributed with a mean of μ=800\mu = 800 days and a standard deviation of σ=50\sigma = 50 days.

(a) The probability that a randomly selected battery lasts less than kk days is 0.150.15. Find the value of kk.

[3]
(b)

(b) A battery is randomly selected. It is known that the battery lasts longer than 750750 days. What is the probability that it lasts longer than 820820 days?

[4]
(c)

(c) A retailer orders 200200 batteries. What is the probability that at most 6262 of them have a lifespan less than 770770 days?

[5]

Question 5

MediumPaper 1 · calculator6 marks
(a)

A beverage company uses an automated machine to fill plastic bottles with water. The actual volume of water, VV, in a randomly chosen bottle is normally distributed with a mean of 500 mL and a standard deviation of 15 mL.

In a quality control check, if a randomly chosen bottle contains less than 485 mL of water, the company fails the check.

Find the probability that the company fails this quality control check.

[2]
(b)

A quality assurance manager suggests that a fairer check would be to pass if the mean volume of eight randomly chosen bottles is greater than 485 mL.

Find the probability of passing the quality control check if the manager's suggestion is followed.

[4]

Question 6

HardPaper 2 · calculator14 marks
(a)

A manufacturer claims that the lifespan of a new batch of LED light bulbs follows a normal distribution with a mean of 50005000 hours and a standard deviation of 10001000 hours. To test this claim, a random sample of 250250 light bulbs was selected, and their lifespans were recorded. The observed frequencies are shown in the table below.

Lifespan (hours)Observed Frequency
x<3500x < 35001010
3500≤x<45003500 \le x < 45005050
4500≤x<55004500 \le x < 55009595
5500≤x<65005500 \le x < 65007070
x≥6500x \ge 65002525

(a) Copy and complete the following table of expected frequencies, assuming the manufacturer's claim is true. Give your answers to two decimal places.

[4]
(b)

Using a χ2\chi^2 distribution at the 5%5\% level of significance, test the hypothesis that the lifespan of the light bulbs follows a normal distribution with mean 50005000 hours and standard deviation 10001000 hours.

You should state the null and alternative hypotheses, clearly show your working for the χ2\chi^2 statistic, and justify your conclusion.

The correct critical value may be selected from the following table, where X(5%)2X^2_{(5\%)} is the value such that P(X>X(5%)2)=0.05P(X > X^2_{(5\%)}) = 0.05.

Degrees of freedomX(5%)2X^2_{(5\%)}
113.843.84
225.995.99
337.827.82
449.499.49
5511.0711.07
[10]

Question 7

MediumPaper 1 · calculator6 marks
(a)

A factory produces two types of electronic components: standard (S) and premium (P). The weight of these components is a critical characteristic for quality control.

The weights of standard components are known to be normally distributed with a mean of 150 grams and a standard deviation of 5 grams.

The weights of premium components are known to be normally distributed with a mean of 165 grams and a standard deviation of 8 grams.

A quality control machine classifies a component as 'premium' if its weight is found to be above 158 grams; otherwise, it is classified as 'standard'.

The factory's quality control manager uses the null hypothesis that, in the absence of other information, a component is standard.

Calculate the probability of making a Type I error when classifying a component.

[2]
(b)

Calculate the probability of making a Type II error when classifying a component.

[2]
(c)

It is known that 80% of the components produced are standard, and 20% are premium.

Calculate the overall probability that a randomly selected component is misclassified by the machine.

[2]

Question 8

HardPaper 2 · calculator12 marks
(a)

(a) A logistics company handles two types of packages: small (S) and large (L). The weight of each type of package follows a normal distribution with parameters as shown in this table:

Type of Package

Mean weight (kg)

Standard deviation (kg)

S (Small)

1.51.5

0.20.2

L (Large)

5.05.0

0.50.5

One package of each type is selected at random. Find the probability that the large package weighs less than four times the weight of the small package.

[5]
(b)

(b) One large package and three small packages are selected at random. Find the probability that the large package weighs more than the total weight of the three small packages.

[7]

Question 9

MediumPaper 1 · calculator5 marks
(a)

The weight of coffee bean packages from a certain supplier is modelled by a normal distribution with a mean of 480480 grams.

A package weighing 504504 grams is two standard deviations from the mean.

Find the standard deviation for the weight of the coffee bean packages.

[2]
(b)

It is found that 75%75\% of these coffee bean packages have weights between pp and qq grams, where p<qp < q. This interval includes packages weighing 498498 grams.

Show that the region of the normal distribution between pp and qq is not symmetrical about the mean.

[3]

Question 10

HardPaper 2 · calculator16 marks
(a)

A factory produces specialized electronic components. The total "quality score" of a component, TT, is a combination of scores from three independent inspection stages:

  • Stage 1: Automated visual inspection. The score XX from this stage has an expectation of 2.52.5 and a standard deviation of 0.70.7.
  • Stage 2: Manual functional test. A batch of 88 critical functions are tested, and the score YY is the number of functions that pass. Each function has a 0.350.35 probability of passing, independently.
  • Stage 3: Environmental stress test. The score ZZ from this stage, representing the number of successful stress cycles, follows a Poisson distribution with a mean of 4.24.2.

The overall quality score for a component is given by T=X+2Y+ZT = X + 2Y + Z.

Calculate the expected value and variance of the total quality score TT.

[6]
(b)

Given that the distribution of TT can be approximated by a Normal distribution, find the probability that a randomly selected component has a total quality score between 1010 and 1414 (inclusive of 1010, exclusive of 1414).

[4]
(c)

The factory manager wants to ensure that the mean total quality score of a sample of nn components is within 0.80.8 units of the true mean, with a probability of at least 0.950.95. Find the minimum sample size nn required.

[6]

Question 11

MediumPaper 1 · calculator7 marks
(a)

A new type of LED bulb has a lifespan, LL, which is normally distributed with a mean of 5000 hours and a standard deviation of 300 hours.

The following curve represents this distribution. It is known that P(L<a)=0.1P(L < a) = 0.1 and P(L>b)=0.1P(L > b) = 0.1.

Normal distribution curve with mean 5000 and standard deviation 300. Areas to the left of 'a' and to the right of 'b' are shaded and labelled 10%.

(a) Calculate the probability that a randomly selected LED bulb lasts more than 4700 hours.

[2]
(b)(i)

(b) (i) Find the value of aa.

[1]
(b)(ii)

(b) (ii) Find the value of bb.

[2]
(c)

(c) Two LED bulbs are selected at random from a large batch. Find the probability that they both have a lifespan less than 4700 hours.

[2]

Question 12

HardPaper 2 · calculator8 marks

A pharmaceutical company manufactures a drug where the concentration of the active ingredient, CC (in mg/mL), is crucial. The concentration is influenced by the reaction time, tt (in minutes), during the production process.

The relationship between the concentration CC and the reaction time tt is given by C=0.5t+10C = 0.5t + 10.

Due to slight variations in the manufacturing process, the reaction time tt is normally distributed with a mean of 2525 minutes and a standard deviation of 1.51.5 minutes. That is, t∼N(25,1.52)t \sim N(25, 1.5^2).

The quality control department assigns points to each batch based on the final concentration CC:

Interval (mg/mL)Points scored
23<C≤24.523 < C \le 24.51515
22<C≤2322 < C \le 2388
21<C≤2221 < C \le 2222
Otherwise00

Find the expected total quality points scored from 200200 batches.

Question 13

MediumPaper 1 · calculator8 marks
(a)

P1: The time, in seconds, a new automated machine takes to produce a standard component is normally distributed with a mean of 120120 seconds and a standard deviation of 88 seconds.

(a) Find the probability that, on a randomly selected run, the machine takes longer than 125125 seconds to produce a component.

[2]
(b)

(b) Find the probability that on a randomly selected run, the machine takes between 115115 and 122122 seconds to produce a component.

[3]
(c)

(c) The factory produces 7575 components in a day.

On how many occasions should the factory expect the machine to take less than 110110 seconds to produce a component?

[3]

Question 14

HardPaper 2 · calculator14 marks
(a)

The lifespan of a new model of LED light bulb, LL, is normally distributed with a mean of 1200012000 hours and a standard deviation of 800800 hours.

Sketch a diagram showing this information.

[2]
(b)

Find the proportion of these LED bulbs that have a lifespan between 1100011000 hours and 1300013000 hours.

[2]
(c)

A large batch of 200200 LED bulbs is purchased. Determine the expected number of bulbs in this batch that will have a lifespan of less than 1080010800 hours.

[3]
(d)

It is observed that 15%15\% of the LED bulbs last longer than hh hours. Estimate the value of hh.

[3]
(e)

Ten of these LED bulbs are chosen at random. Find the probability that exactly two of them last longer than 1280012800 hours.

[4]

Question 15

MediumPaper 1 · calculator5 marks
(a)

The volume of liquid in bottles produced by a beverage company is normally distributed with a mean of 750750 mL and a standard deviation of 55 mL. A bottle is considered 'underfilled' if its volume is less than 740740 mL.

(a) Find the percentage of bottles that are classified as underfilled.

[2]
(b)

Bottles are packed into crates, with each crate containing 1212 bottles.

(b) Find the probability that a random crate has at most two underfilled bottles.

[3]

Question 16

HardPaper 2 · calculator14 marks
(a)(i)

The diameter of a certain type of industrial component is modelled by a normal distribution with a mean of 5050 mm and a standard deviation of 0.50.5 mm.

Find the probability that a randomly selected component has a diameter less than 50.750.7 mm.

[2]
(a)(ii)

Find the probability that a randomly selected component has a diameter greater than 51.251.2 mm.

[1]
(b)

Assuming the diameters of components are independent, find the probability that two consecutive components both have a diameter greater than 51.251.2 mm.

[2]
(c)

A component is classified as 'premium' if its diameter is between 49.549.5 mm and 50.550.5 mm. A batch of three components is considered 'high quality' if all three components in the batch are premium. Find the probability that a randomly selected batch is NOT high quality.

[2]
(d)(i)

In a production run, 1212 batches of three components are produced. Find the probability that at least 77 of these batches are high quality.

[3]
(d)(ii)

Find the probability that between 77 and 1010 (exclusive of 1010) of these batches are high quality.

[2]
(d)(iii)

Given that at least 77 batches are high quality, find the probability that less than 1010 batches are high quality.

[2]

Question 17

MediumPaper 2 · calculator13 marks
(a)(i)

A factory produces light bulbs. The probability that a randomly selected light bulb is defective is p=0.15p = 0.15. For a single light bulb, let the discrete random variable XX equal 11 if the bulb is defective, and 00 if it is not defective.

(a) For the random variable XX, write down

(i) the mean

[1]
(a)(ii)

(ii) the variance.

[1]
(b)(i)

A quality control manager inspects a random sample of n=120n = 120 light bulbs. Let the sample mean Xˉ=1n∑i=1120Xi\bar{X} = \frac{1}{n} \sum_{i=1}^{120} X_i be the proportion of defective bulbs in the sample. The Central Limit Theorem states that the distribution of Xˉ\bar{X} can be approximated by a normal distribution for a sufficiently large nn.

(b) Write down the

(i) mean

[1]
(b)(ii)

(ii) variance

of Xˉ\bar{X}, when approximated by a normal distribution.

[1]
(c)

(c) Use this normal approximation to estimate P(Xˉ>0.18)P(\bar{X} > 0.18).

[3]
(d)

(d) Hence, write down P(∑i=1120Xi>k)P\left(\sum_{i=1}^{120} X_i > k\right) for an appropriate value of kk, when using this normal approximation.

[2]
(e)(i)

(e) Let T=∑i=1120XiT = \sum_{i=1}^{120} X_i be the total number of defective light bulbs in the sample.

(i) State the true distribution that TT satisfies.

[1]
(e)(ii)

(ii) Find the exact value of P(T>21.5)P(T > 21.5), which can be construed as the probability that more than 2121 bulbs are defective.

[3]

Question 18

HardPaper 2 · calculator21 marks
(a)(i)

The lifespans, tt, of 250 LED light bulbs are recorded in the following table.

Lifespan (hours)Frequency
0≤t<10000 \le t < 100020
1000≤t<15001000 \le t < 150060
1500≤t<20001500 \le t < 200090
2000≤t<25002000 \le t < 250055
2500≤t<30002500 \le t < 300025

This table is used to create a cumulative frequency graph.

Write down the mid-interval value of the class 0≤t<10000 \le t < 1000.

[1]
(a)(ii)

Calculate an estimate of the mean lifespan of the 250 light bulbs.

[3]
(b)

Use the cumulative frequency curve (which would be provided in an exam) to estimate the interquartile range. Assume the lower quartile (Q1Q_1) is 13001300 hours and the upper quartile (Q3Q_3) is 21502150 hours.

[3]
(c)

A light bulb from the data set had a lifespan of 34003400 hours.

Use your answer to part (b) to estimate whether this light bulb's lifespan is an outlier for this data. Justify your answer.

[3]
(d)

It is believed that the lifespans of these LED light bulbs follow a normal distribution with mean 17401740 hours and standard deviation 450450 hours.

It is decided to perform a χ2\chi^2 goodness of fit test on the data to determine whether this sample of 250 light bulbs could have plausibly been drawn from an underlying distribution N(1740,4502)N(1740, 450^2).

Write down the null and the alternative hypotheses for the test.

[2]
(e)(i)

As part of the test, the following table is created.

Lifespan of light bulb (hours)Observed frequencyExpected frequency
t<1000t < 10002014.0
1000≤t<15001000 \le t < 15006060.1
1500≤t<20001500 \le t < 200090a
2000≤t<25002000 \le t < 25005560.1
t≥2500t \ge 250025b

Find the value of aa and the value of bb. Give your answers to one decimal place.

[5]
(e)(ii)

Hence, perform the test to a 5% significance level, clearly stating the conclusion in context.

[4]

Question 19

MediumPaper 1 · calculator17 marks
(a)

(a) Assume that the volume of liquid in bottles produced by a certain factory follows a normal distribution with a mean of 10051005 ml and a standard deviation of 88 ml.

Find the probability that a randomly chosen bottle contains more than 10101010 ml.

[3]
(b)

(b) Find the interquartile range of the liquid volume in the bottles.

[4]
(c)

(c) Find the volume, in ml, that is exceeded by 15%15\% of the bottles.

[3]
(d)

(d) A quality control inspector randomly selects 1212 bottles. Find the probability that at most 44 of them contain more than 10101010 ml.

[4]
(e)

(e) In a large batch of 500500 bottles, estimate how many would contain less than 995995 ml.

[3]

Question 20

HardPaper 2 · calculator25 marks
(a)(i)

The lifespan of a new type of LED bulb, in hours, can be modelled by a normal distribution with a mean of 1200012000 hours and a standard deviation of 800800 hours.

A randomly selected LED bulb is chosen.

(a) Calculate the probability that its lifespan is

(i) less than 1100011000 hours.

[4]
(a)(ii)

(ii) between 1150011500 hours and 1250012500 hours.

[4]
(b)

(b) 15%15\% of LED bulbs have a lifespan of more than hh hours.

Calculate the value of hh.

[2]
(c)(i)

A manufacturer wants to determine if a sample of 250250 LED bulbs from a new production batch could have been chosen from a normally distributed population with a mean of 1200012000 hours and a standard deviation of 800800 hours.

They perform a χ2\chi^2 goodness of fit test at the 5%5\% significance level. They begin by creating the following frequency table:

Lifespan, hh (hours)Observed frequencyExpected frequency
h≤11000h \le 11000151526.41226.412
11000<h≤1200011000 < h \le 12000100100a
12000<h≤1300012000 < h \le 13000110110b
h>13000h > 13000252526.41226.412

(c) Calculate, correct to four significant figures, the value of

(i) a.

[2]
(c)(ii)

(ii) b.

[2]
(d)

The hypotheses for the manufacturer's test are:

H0H_0: The lifespans of the LED bulbs are drawn from a normally distributed population with mean 1200012000 hours and standard deviation 800800 hours.

H1H_1: The lifespans of the LED bulbs are not drawn from a normally distributed population with mean 1200012000 hours and standard deviation 800800 hours.

(d) Write down the degrees of freedom for this test.

The critical value for this test is 7.8157.815.

[1]
(e)

(e) Perform the χ2\chi^2 goodness of fit test and state your conclusion, justifying your reasoning.

[4]
(f)

A competitor claims that their new 'Brand A' LED bulbs last longer on average than the manufacturer's 'Brand B' LED bulbs.

Random samples of 1212 Brand A bulbs and 1010 Brand B bulbs are chosen, and their lifespans (in hours) are measured:

Lifespans of Brand A bulbs (hours):

125001250011800118001320013200121001210012900129001190011900127001270012300123001310013100120001200012600126001220012200

Lifespans of Brand B bulbs (hours):

1210012100115001150012800128001190011900124001240011700117001200012000122001220011600116001230012300

The competitor performs a t-test at the 5%5\% significance level. It is assumed that the populations are normally distributed and have equal variances.

(f) Write down the null and alternative hypotheses for this test.

[2]
(g)

(g) Perform the t-test and state the conclusion, justifying your reasoning.

[4]

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What does Normal distribution + bell curve (+inv normal) cover in IB Maths AI?

Normal distribution is a continuous probability distribution. Notation: X sim N(μ, σ^2), where μ is the population mean and σ^2 is the population variance. The distribution is symmetrical, bell-shaped, and has a single mode.

Is Normal distribution + bell curve (+inv normal) SL or HL?

Both. SL and HL students study Normal distribution + bell curve (+inv normal) to the same depth.

How do I revise Normal distribution + bell curve (+inv normal) for IB Maths AI?

Start from the core idea: normal distribution is a continuous probability distribution. In the exam: the exclusion of z-scores is the single most useful fact here, and it is the biggest difference from AA. Everything is done on the GDC with μ and σ given. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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