Binomial and normal distributions 201-SN1-RE: revision sheet

Q: How do you know whether a variable follows a binomial distribution?

Check four conditions in the problem: a number of trials fixed in advance, two outcomes per trial, the same probability of success on every trial, and independent trials. Drawing without replacement is still acceptable if the sample is at most 5% of the population. A common cause, such as the same tray of potting soil for all the seeds, destroys independence even if the probability is the same.

Q: When can a binomial distribution be approximated by a normal distribution?

When n times p and n times 1 minus p are both at least 5, according to the most common criterion; some CEGEPs require 10. You then take a normal with mean np and standard deviation the square root of np(1 - p). A large n is not enough: with p equal to 0.99 and n equal to 200, n times 1 minus p is 2 and the approximation is refused.

Q: Why do we add or subtract 0.5 in the continuity correction?

Because the binomial counts integers and the normal measures intervals. The bar of the integer k occupies the interval from k minus 0.5 to k plus 0.5. For “at least k”, the area therefore starts at k minus 0.5; for “at most k”, it stops at k plus 0.5. You always take the outer edge of the bar at the boundary of the event.

Q: How do you read the normal table with a negative z-score?

The table gives the area to the left of z. For a negative score, use the symmetry of the curve: the area to the left of minus a equals the area to the right of a, that is 1 minus the area to the left of a. Quick check: the area to the right of a negative score contains the peak, so it is more than 0.5.

Q: Why is the probability of an exact value zero for a normal distribution?

For a continuous variable, a probability is an area under the density, and the area above a single point is zero. This does not mean that the value is impossible: it is being asked for with infinitely many decimal places. A real measurement is rounded, so it is an interval, and it is that interval that has a probability.

The thread of the chapter

A binomial COUNTS, a normal MEASURES: the binomial gives a probability to each integer, so $<$ and $\le$ change the answer; the normal gives probability only to an interval, and $P(X=a)=0$ . Going from one to the other turns each integer into an interval one unit wide: that is the whole of the continuity correction.

Topics covered

Bernoulli trial
Binomial distribution
Expected value and variance of a binomial
Normal distribution
Standard normal distribution
Standardization and z-score
Standard normal table
Normal approximation to the binomial
Continuity correction
Poisson distribution

The chapter at a glance

Binomial, normal, approximation: recognizing the distribution

First read what the variable does: it counts successes, or it measures a quantity. Everything else follows.

Criterion	Binomial it counts	Normal it measures	Approximation we count, we compute by measuring
What $X$ does	Number of successes in $n$ trials	Continuous quantity: mass, pH, length	A binomial count with many trials
Conditions	$n$ fixed, two outcomes, same $p$ , independence	Model assumed or given, $\mu$ and $\sigma$ known	$np\ge 5$ and $n(1-p)\ge 5$
Parameters	$n$ and $p$ ; $E=np$ , $V=np(1-p)$	$\mu$ and $\sigma$ (or $\sigma^{2}$ , depending on the notation)	$\mu=np$ , $\sigma=\sqrt{np(1-p)}$
$P(X=a)$	$\binom{n}{a}p^{a}(1-p)^{n-a}>0$	$0$ : a single value has no area	Area from $a-0.5$ to $a+0.5$
$<$ versus $\le$	Different: $P(X<k)=P(X\le k-1)$	Identical	They decide the bound: $k\pm 0.5$
Calculation	Formula, sum or complement	$z=\dfrac{x-\mu}{\sigma}$ then the table	Continuity correction, then the table

Binomial it counts

What $X$ does: Number of successes in $n$ trials
Conditions: $n$ fixed, two outcomes, same $p$ , independence
Parameters: $n$ and $p$ ; $E=np$ , $V=np(1-p)$
$P(X=a)$: $\binom{n}{a}p^{a}(1-p)^{n-a}>0$
$<$ versus $\le$: Different: $P(X<k)=P(X\le k-1)$
Calculation: Formula, sum or complement

Normal it measures

What $X$ does: Continuous quantity: mass, pH, length
Conditions: Model assumed or given, $\mu$ and $\sigma$ known
Parameters: $\mu$ and $\sigma$ (or $\sigma^{2}$ , depending on the notation)
$P(X=a)$: $0$ : a single value has no area
$<$ versus $\le$: Identical
Calculation: $z=\dfrac{x-\mu}{\sigma}$ then the table

Approximation we count, we compute by measuring

What $X$ does: A binomial count with many trials
Conditions: $np\ge 5$ and $n(1-p)\ge 5$
Parameters: $\mu=np$ , $\sigma=\sqrt{np(1-p)}$
$P(X=a)$: Area from $a-0.5$ to $a+0.5$
$<$ versus $\le$: They decide the bound: $k\pm 0.5$
Calculation: Continuity correction, then the table

Key: the words of the table

Successes: The outcome counted in each trial: the seed germinates, the plant recovers, the rain event is very acidic.
Continuous: A variable that can take every value in an interval; a probability is then an area under the density.
Independence: The result of one trial does not change the probability of the others; a shared tray of soil or a contagion destroys it.
And: Both conditions are required; with p close to 1, it is n(1 - p) that fails.
Different: For a variable that counts, the gap between P(X < k) and P(X ≤ k) is exactly P(X = k).
Complement: Computing 1 - P(opposite) when the opposite has fewer terms: at least one, at least 9 out of 10.
Continuity correction: The shift of half a unit that turns the bar of an integer k into the interval from k - 0.5 to k + 0.5.

The misleading case: a binomial with a large $n$ whose $p$ is close to 0 or to 1. It is still a count, and the normal approximation is refused as long as $np$ or $n(1-p)$ is below 5.

The essentials

Counting or measuring

• A BINOMIAL variable counts: each integer from $0$ to $n$ has its probability, the height of a bar.
• A NORMAL variable measures: a probability is an AREA under the density, and $P(X=a)=0$ for every value $a$ .
• Consequence: $P(X<8)$ and $P(X\le 8)$ differ by $P(X=8)$ for a binomial, and are equal for a normal.
• A measured and rounded value is also an interval: a wing displayed as $9.3$ mm has a length in $[9.25, 9.35)$ .

On the left, the binomial

\mathcal{B}(12,\,0.3)

: the bar of 4 carries

0.231

. On the right, a normal: the green area is a probability, the red line at

23

carries none.

Writing “ $X$ is discrete, so $P(X<8)=P(X\le 7)$ ” before computing avoids half of the bound errors.

The four conditions of the binomial

• 1) a number $n$ of trials FIXED in advance;
• 2) two outcomes per trial, success or failure;
• 3) the SAME probability of success $p$ on every trial;
• 4) INDEPENDENT trials.
• Without replacement, keep the binomial if the sample is at most $5\%$ of the population. Beyond that, count directly: 6 seeds drawn from a packet of 20 with 17 viable give $\frac{\binom{17}{6}}{\binom{20}{6}}\approx 0.3193$ , not $(0.85)^{6}\approx 0.3771$ .

The conditions are CHECKED against the situation, one by one, with the words of the problem that justify them (“one pot per seed”, “10 seeds out of 40,000”).

The bar of k is an interval

• To approximate $\mathcal{B}(n,\,p)$ , take the normal with mean $\mu=np$ and standard deviation $\sigma=\sqrt{np(1-p)}$ , if $np\ge 5$ and $n(1-p)\ge 5$ .
• The bars have width 1 and are centred on the integers: the bar of $k$ covers $[k-0.5, k+0.5]$ .
• $P(X=k)\approx P(k-0.5\le Y\le k+0.5)$ ; $P(X\le k)\approx P(Y\le k+0.5)$ ; $P(X\ge k)\approx P(Y\ge k-0.5)$ .

The orange bar of

12

0.1201

; the green area under the curve, from

11.5

12.5

, is

0.1200

: the continuity correction is this half unit on each side.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Applying the binomial to a draw without replacement from a small population

the whole question, and often the next one

What not to write

“ $85\%$ of the seeds in the packet are viable, so the number of seeds out of 6 that germinate follows $\mathcal{B}(6,\,0.85)$ .”

What to write

“6 seeds out of 20 are $30\%$ of the packet, more than $5\%$ : $p$ changes with each draw. By counting, $P(\text{all 6 germinate})=\frac{\binom{17}{6}}{\binom{20}{6}}\approx 0.3193$ .”

Why: The proportion is right, the model is wrong: after one viable seed is drawn, 16 out of 19 remain, and $\frac{16}{19}\ne 0.85$ .

2. Believing that the same probability of success is enough

1 to 2 marks for justifying the model

What not to write

“Each seed in the tray germinates with probability $0.85$ , so $P(\text{all 10 germinate})=(0.85)^{10}\approx 0.1969$ .”

What to write

“The seeds share soil that may be contaminated: they are not independent. Condition on the tray: $0.1(0.40)^{10}+0.9(0.90)^{10}\approx 0.3138$ .”

Why: A common cause makes successes come together and failures come together; you need a physical reason to believe in independence, such as one pot per seed.

3. Taking the variance for the standard deviation

the whole rest of the calculation, about 2 marks

What not to write

“ $X\sim\mathcal{B}(200,\,0.85)$ , so $\sigma=np(1-p)=25.5$ .”

What to write

“ $V(X)=np(1-p)=25.5$ , so $\sigma=\sqrt{25.5}\approx 5.05$ .”

Why: A z-score divides by a standard deviation, in the unit of $X$ . With $25.5$ , every z-score is five times too small and every probability is wrong.

4. Confusing “fewer than 8” and “at most 8” for a binomial

1 mark per question

What not to write

“ $P(X<8)=P(X\le 8)\approx 0.4557$ ” for $X\sim\mathcal{B}(10,\,0.85)$ .

What to write

“ $X$ is discrete: $P(X<8)=P(X\le 7)\approx 0.1798$ , and $P(X\le 8)\approx 0.4557$ ; the difference is $P(X=8)\approx 0.2759$ .”

Why: A variable that counts gives a probability to each integer; the sign decides whether the bar of 8 is in the calculation.

5. Reading the table on the wrong side for a negative z

1 mark, and the conclusion of the exercise

What not to write

“ $P(Z>-0.62)=1-\Phi(0.62)=1-0.7324=0.2676$ .”

What to write

“By symmetry, $P(Z>-0.62)=P(Z<0.62)=\Phi(0.62)=0.7324$ .”

The two green areas are mirror images of each other: to the right of

-0.62

and to the left of

0.62

, both

0.7324

, more than half.

Why: The area to the right of a negative number contains the whole right half of the curve: it is more than $0.5$ . A quick sketch of the bell shows it in two seconds.

6. Dividing by the variance when the problem gives it

the whole question

What not to write

“ $X\sim\mathcal{N}(250,\,9)$ , so $z=\dfrac{256-250}{9}\approx 0.67$ .”

What to write

“In the notation $\mathcal{N}(\mu,\,\sigma^{2})$ , $9$ is the variance: $\sigma=3$ and $z=\dfrac{256-250}{3}=2.00$ .”

Why: The substitution $x=\mu+\sigma z$ brings $\sigma$ out of $dx$ , never $\sigma^{2}$ . Read the notation of your textbook: some write the standard deviation, others the variance.

7. Putting the half unit on the wrong side

1 to 2 marks, half of the question

What not to write

“ $P(X\ge 175)\approx P(Y\ge 175.5)=0.1379$ ” for $X\sim\mathcal{B}(200,\,0.85)$ .

What to write

“The bar of 175 starts at $174.5$ : $P(X\ge 175)\approx P(Y\ge 174.5)=0.1867$ , for $0.1876$ exactly.”

Why: Starting at $175.5$ removes the whole bar of 175, $P(X=175)\approx 0.05$ : you are actually computing $P(X>175)$ .

8. Approximating by a normal as soon as n is large

the whole question

What not to write

“ $n=200$ is large, so for $p=0.99$ the binomial is approximately normal.”

What to write

“ $n(1-p)=2<5$ : the approximation is refused. The normal would give $0.0375$ to “more than 200 germinated seeds out of 200”, an impossible event.”

Why: A binomial pressed against its bound $0$ or $n$ is very skewed; a symmetric curve describes it poorly. The threshold “ $n\ge 30$ ” belongs to the central limit theorem, a different result.

9. Forgetting the sign of z in a reverse reading

the whole question, the system no longer makes sense

What not to write

“ $10\%$ of the tablets weigh less than $246$ mg, so $\mu+1.28\,\sigma=246$ .”

What to write

“The bound is below the mean: $\Phi(z)=0.10$ gives $z\approx -1.28$ , so $\mu-1.28\,\sigma=246$ .”

Why: A value below the mean always has a negative z-score; a sketch of the bell with the small area on the left is a reminder.

Which method to choose

Which distribution, which calculation

Look at what the problem does with the variable: does it count it, measure it, or ask for a value from a percentage?

If we count successes in $n$ fixed, independent trials with the same $p$ , with few terms to add → exact binomial, by the formula or by the complement

Example: $P(X\ge 9)$ for $\mathcal{B}(10,\,0.85)$ : $0.3474+0.1969=0.5443$
If we count over a large $n$ , with $np\ge 5$ and $n(1-p)\ge 5$ → normal approximation, with the continuity correction

Example: $\mathcal{B}(150,\,0.4)$ : $P(X\le 45)\approx\Phi(-2.42)=0.0078$
If we count over a large $n$ , but $np<5$ or $n(1-p)<5$ → exact binomial (calculator), no normal

Example: $\mathcal{B}(40,\,0.0436)$ : $P(Y\ge 4)\approx 0.0955$
If we measure a continuous quantity with a normal distribution → standardize $z=\dfrac{x-\mu}{\sigma}$ , then read the table

Example: mass $\mu=250$ , $\sigma=3$ : $P(X>251.8)=1-\Phi(0.60)=0.2743$
If a percentage is given and a value is wanted → reverse reading: $z$ in the margin, then $x=\mu+z\sigma$

Example: the heaviest $1\%$ : $250+2.33\times 3\approx 257.0$ mg
If we count rare events in a volume or a time interval, with no number of trials → Poisson distribution with parameter $\lambda$ (beyond the minimum of the course framework)

Example: $\lambda=4$ : $P(X\le 2)=13\,e^{-4}\approx 0.2381$

If no branch applies, go back to the four conditions: it is almost always independence or the fixed number of trials that is missing.

Which bound for the continuity correction

First translate the event into integers, then take the OUTER edge of the bar at the boundary of the event.

For

P(X\ge 175)

, the area starts at

174.5

and covers the whole bar of 175 (green); starting at

175.5

removes it (red).

If $X\le k$ (at most $k$ ) → $Y\le k+0.5$

Example: $X\le 20$ : $Y\le 20.5$
If $X<k$ (fewer than $k$ ) → $X\le k-1$ , so $Y\le k-0.5$

Example: $X<8$ : $Y\le 7.5$
If $X\ge k$ (at least $k$ ) → $Y\ge k-0.5$

Example: $X\ge 175$ : $Y\ge 174.5$
If $X>k$ (more than $k$ ) → $X\ge k+1$ , so $Y\ge k+0.5$

Example: $X>175$ : $Y\ge 175.5$
If $X=k$ (exactly $k$ ) → $k-0.5\le Y\le k+0.5$

Example: $X=170$ : $0.0796$

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a normal approximation

When to use it: The problem asks for a binomial probability with many terms, or says “by the normal approximation”.

1 Name the exact distribution and justify its four conditions with the words of the problem.
2 Check $np\ge 5$ and $n(1-p)\ge 5$ by writing out both numbers, and name the criterion of the course.
3 Give $\mu=np$ and $\sigma=\sqrt{np(1-p)}$ , with the square root.
4 Translate the event into integers, then write the continuity correction: $P(X\ge k)\approx P(Y\ge k-0.5)$ .
5 Standardize, round $z$ to two decimal places, read the table, and conclude with a sentence in context.

Concluding sentence

“Since $np=60\ge 5$ and $n(1-p)=90\ge 5$ , we approximate $X$ by a normal $Y$ with mean $60$ and standard deviation $6$ : $P(X\le 45)\approx P(Y\le 45.5)=\Phi(-2.42)=0.0078$ .”

The trap: Skipping the check of the conditions: an approximation used without checking is marked wrong, even if the number is right.

Marking: 1 mark for the model, 1 mark for the conditions, 1 mark for the correction, 1 mark for the calculation and the sentence

Proving E(X) = np

When to use it: The problem asks you to prove the expected value or the variance of a binomial.

1 Let $G(t)=(q+pt)^{n}$ and expand it by the binomial theorem: $G(t)=\sum_{k=0}^{n}P(X=k)\,t^{k}$ .
2 Differentiate both expressions: $np(q+pt)^{n-1}=\sum_{k=1}^{n}k\,P(X=k)\,t^{k-1}$ .
3 Evaluate at $t=1$ , where $q+p=1$ : $E(X)=np$ .
4 For the variance, differentiate again: $G''(1)=E\big(X(X-1)\big)=n(n-1)p^{2}$ , then $V(X)=E(X^{2})-E(X)^{2}=np(1-p)$ .

Concluding sentence

“At $t=1$ , $G'(1)=\sum_{k=0}^{n}k\,P(X=k)=E(X)$ and $G'(1)=np(q+p)^{n-1}=np$ , so $E(X)=np$ .”

The trap: Taking $G''(1)$ for $E(X^{2})$ : the second derivative brings down $k(k-1)$ , not $k^{2}$ .

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Which side of the bell

Sketch the bell in three seconds and shade the requested region: an area that contains the peak is more than $0.5$ , a tail is less than $0.5$ .

Finding $0.2676$ for $P(Z>-0.62)$ is impossible: the region contains the peak, the answer is $0.7324$ .
The calculator confirms

TI: binomcdf(n, p, k) gives $P(X\le k)$ , normalcdf(a, b, μ, σ) an area, invNorm(area, μ, σ) a reverse reading. Casio: Bcd, Ncd and InvN in the distribution menu.

binomcdf(200, 0.85, 174) $\approx 0.8124$ , so $P(X\ge 175)\approx 0.1876$ : the approximation $0.1867$ is confirmed; $0.1611$ would give away the forgotten correction.
The order of magnitude from standard deviations

Place the bound in standard deviations around $\mu$ : at $1\sigma$ , about $16\%$ beyond; at $2\sigma$ , about $2.3\%$ ; at $3\sigma$ , less than $0.2\%$ .

45 infected plants when $\mu=60$ and $\sigma=6$ : $2.5\sigma$ below the mean, so a probability of the order of $1\%$ . Finding $0.2$ would signal an error.
The exact distribution does not go past its bounds

A binomial lives between $0$ and $n$ . If the approximating normal puts a visible area beyond, the approximation is refused.

$\mathcal{B}(200,\,0.99)$ : the normal gives $0.0375$ beyond 200, the signal that $n(1-p)=2$ is too small.

The typical problem, taken apart

A germination test: at most 85 seeds out of 120?

A lot of clover seeds advertises a germination rate of $75\%$ . 120 seeds taken at random from a large lot are sown, one per cell, and $X$ denotes the number of seeds that germinate.

Compute an approximate value of $P(X\le 85)$ .

Step 1

The 120 seeds come from a large lot, each in its own cell: $n=120$ is fixed, each seed germinates or not with the same probability $p=0.75$ , independently of the others. So $X\sim\mathcal{B}(120,\,0.75)$ .

Why

The model is justified by the words of the problem: “large lot” for the 5% rule, “one per cell” for independence. That is the first mark on the marking scheme.

Step 2

$np=90\ge 5$ and $n(1-p)=30\ge 5$ : the normal approximation is accepted, with $\mu=90$ and $\sigma=\sqrt{22.5}\approx 4.743$ .

Why

Both numbers are written out, not just “the conditions are met”. The square root is a reminder that we divide by a standard deviation.

Step 3

“At most 85” contains the bar of 85, which ends at $85.5$ : $P(X\le 85)\approx P(Y\le 85.5)$ .

Why

This is the step that separates the binomial from the normal: a count is turned into a measured interval, through the outer edge of the bar at the boundary of the event.

Step 4

$z=\dfrac{85.5-90}{4.743}\approx -0.95$ , then $\Phi(-0.95)=0.1711$ .

Why

Round $z$ to two decimal places BEFORE using the table; the negative sign says that the area is a left tail, below $0.5$ .

Step 5

Check: binomcdf(120, 0.75, 85) $\approx 0.1709$ . Without the correction, we would get $\Phi(-1.05)=0.1469$ .

Why

The calculator check confirms the approximation and shows what the forgotten half unit costs: 0.024, that is 14% of the answer.

The conclusion, written out

“About $17\%$ of trials of 120 seeds would give at most 85 germinations if the advertised rate of $75\%$ is correct: such a result would be nothing unusual.”

The classic mistake on this problem: Computing $P(Y\le 85)$ without the correction, or standardizing with $\sigma^{2}=22.5$ : you get $0.1469$ or a z-score of $-0.20$ , and the conclusion changes.

Learn by heart

• BINOMIAL: $n$ fixed, two outcomes, same $p$ , independence; $P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$ , $E=np$ , $V=np(1-p)$ .
• Without replacement: the binomial is acceptable if the sample is at most $5\%$ of the population.
• A binomial COUNTS: $P(X<k)=P(X\le k-1)$ . A normal MEASURES: $P(X=a)=0$ .
• STANDARDIZE: $z=\dfrac{x-\mu}{\sigma}$ , table of $\Phi(z)=P(Z\le z)$ , $\Phi(-z)=1-\Phi(z)$ ; reverse reading $x=\mu+z\sigma$ .
• Landmarks: $\Phi(1.28)\approx 0.90$ , $\Phi(1.645)\approx 0.95$ , $\Phi(1.96)=0.975$ , $\Phi(2.33)\approx 0.99$ .
• APPROXIMATION if $np\ge 5$ AND $n(1-p)\ge 5$ : $\mu=np$ , $\sigma=\sqrt{np(1-p)}$ , outer edge of the boundary bar: $k+0.5$ for “at most”, $k-0.5$ for “at least”.

Frequently asked questions

How do you know whether a variable follows a binomial distribution?

Check four conditions in the problem: a number of trials fixed in advance, two outcomes per trial, the same probability of success on every trial, and independent trials. Drawing without replacement is still acceptable if the sample is at most 5% of the population. A common cause, such as the same tray of potting soil for all the seeds, destroys independence even if the probability is the same.

When can a binomial distribution be approximated by a normal distribution?

When n times p and n times 1 minus p are both at least 5, according to the most common criterion; some CEGEPs require 10. You then take a normal with mean np and standard deviation the square root of np(1 - p). A large n is not enough: with p equal to 0.99 and n equal to 200, n times 1 minus p is 2 and the approximation is refused.

Why do we add or subtract 0.5 in the continuity correction?

Because the binomial counts integers and the normal measures intervals. The bar of the integer k occupies the interval from k minus 0.5 to k plus 0.5. For “at least k”, the area therefore starts at k minus 0.5; for “at most k”, it stops at k plus 0.5. You always take the outer edge of the bar at the boundary of the event.

How do you read the normal table with a negative z-score?

The table gives the area to the left of z. For a negative score, use the symmetry of the curve: the area to the left of minus a equals the area to the right of a, that is 1 minus the area to the left of a. Quick check: the area to the right of a negative score contains the peak, so it is more than 0.5.

Why is the probability of an exact value zero for a normal distribution?

For a continuous variable, a probability is an area under the density, and the area above a single point is zero. This does not mean that the value is impossible: it is being asked for with infinitely many decimal places. A real measurement is rounded, so it is an interval, and it is that interval that has a probability.

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Corrected exercises: Binomial and normal distributions, 201-SN1-RE

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11 corrected exercises
110 points
150 minutes

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Revision sheet: binomial and normal distributions (201-SN1-RE)

Binomial, normal, approximation: recognizing the distribution

Key: the words of the table

The essentials

Counting or measuring

The four conditions of the binomial

The bar of k is an interval

The mistakes that cost marks

1. Applying the binomial to a draw without replacement from a small population

2. Believing that the same probability of success is enough

3. Taking the variance for the standard deviation

4. Confusing “fewer than 8” and “at most 8” for a binomial

5. Reading the table on the wrong side for a negative z

6. Dividing by the variance when the problem gives it

7. Putting the half unit on the wrong side

8. Approximating by a normal as soon as n is large

9. Forgetting the sign of z in a reverse reading

Which method to choose

Which distribution, which calculation

Which bound for the continuity correction

How the answer is expected to be written

Writing up a normal approximation

Proving E(X) = np

Check before you hand in

The typical problem, taken apart

A germination test: at most 85 seeds out of 120?

Learn by heart

Frequently asked questions

How do you know whether a variable follows a binomial distribution?

When can a binomial distribution be approximated by a normal distribution?

Why do we add or subtract 0.5 in the continuity correction?

How do you read the normal table with a negative z-score?

Why is the probability of an exact value zero for a normal distribution?

Corrected exercises: Binomial and normal distributions, 201-SN1-RE

This chapter comes back in

See also

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