True or False? A situation with more than two possible outcomes can never be modelled by a binomial distribution.

False. What matters is how the trial is defined, not how many outcomes the situation as a whole has. Counting the yellow cars in a car park is binomial even though cars come in many colours, because each car either is yellow or is not, which gives exactly two outcomes per trial; the other three conditions still have to be checked separately.

A LevelMathsOCRMaths AStatisticsFlashcardsStatistical DistributionsBinomial Distribution

Binomial Distribution (OCR A Level Maths A: Statistics): Flashcards

Exam code: H240

1/13

0Still learning

Know0

FrontThe Binomial Distribution
What four conditions must be met for a variable to follow a binomial distribution?
BackThe Binomial Distribution
The variable must count the number of successes in an experiment where all four of these hold:
there is a fixed, finite number of trials, $n$
the outcome of each trial is independent of the outcomes of the others
each trial has exactly two outcomes, success or failure
the probability of success, $p$ , is constant
If any one of them fails, the variable is not binomial.
BackThe Binomial Distribution
Sign up to unlock flashcards
Join for free to unlock a full flashcard set, track what you know,
and turn revision into real progress.

Was this flashcard helpful?

Cards in this collection (13)

What four conditions must be met for a variable to follow a binomial distribution?
The variable must count the number of successes in an experiment where all four of these hold:
- there is a fixed, finite number of trials, $n$
- the outcome of each trial is independent of the outcomes of the others
- each trial has exactly two outcomes, success or failure
- the probability of success, $p$ , is constant
If any one of them fails, the variable is not binomial.
A variable following a binomial distribution is written $X ~ B (n, p)$ . Complete what each parameter stands for:
$n$ is the number of $______$ and $p$ is the probability of $______$ on each one of them.
The completed sentence is:
$n$ is the number of trials and $p$ is the probability of success on each one of them.
The probability of failure is $1 - p$ , which is often written $q$ .
Five sweets are eaten from a bag containing 6 caramels and 4 marshmallows, and $X$ is the number of caramels eaten.
Which binomial condition does this break, and why?
It breaks independence.
Eating a caramel leaves fewer caramels in the bag, so the chance that the next sweet is a caramel depends on what has already been taken, which also means $p$ is not constant.
Independence is the condition worth checking first, because it is usually the one that fails.
In the binomial probability formula, what is the binomial coefficient $(\binom{n}{x})$ counting?
It counts how many different ways $x$ successes can occur among the $n$ trials.
Think of a probability tree with $n$ trials: $p^{x} {(1 - p)}^{n - x}$ is the probability along one path that has $x$ successes and $n - x$ failures, and every such path carries that same probability.
The coefficient is the number of those paths, so multiplying by it adds them all together.
True or False?
A situation with more than two possible outcomes can never be modelled by a binomial distribution.
False.
What matters is how the trial is defined, not how many outcomes the situation as a whole has.
Counting the yellow cars in a car park is binomial even though cars come in many colours, because each car either is yellow or is not, which gives exactly two outcomes per trial; the other three conditions still have to be checked separately.
How is the binomial distribution $B (n, p)$ related to the expansion of ${(p + q)}^{n}$ when $q$ is the probability of failure?
Every term of the expansion is one of the probabilities.
Expanding ${(p + q)}^{n}$ gives terms $(\binom{n}{x}) p^{x} q^{n - x}$ , which is exactly $P (X = x)$ , so the term containing $p^{x}$ is the probability of getting $x$ successes.
Since $p + q = 1$ the expansion adds to 1, which is why the probabilities of all the possible outcomes total 1.
A random sample of 30 people is taken from a large city in which 30% have blue eyes, and $X$ is the number in the sample with blue eyes.
Why can $X$ be modelled as binomial even though nobody is put back?
Because the population is large, so taking a few people out barely changes the proportion left behind, and $p$ stays close to 0.3 for every person in the sample.
The sample also has to be random, so that each person really does have that same chance of being chosen.
Strictly the trials are not independent, but with a large population the model is close enough to be useful.
How does the shape of the vertical line graph of a binomial distribution change as $p$ changes?
It is symmetrical when $p = 0.5$ , and roughly symmetrical when $p$ is close to 0.5.
When $p$ is close to 0, most of the probability sits at the low values and the graph has a tail to the right; when $p$ is close to 1 it has a tail to the left.
Why can every strict inequality about a binomial variable be rewritten as a weak one?
Because a binomial variable can only take the whole-number values $0, 1, 2, \dots, n$ , so there is nothing lying strictly between two consecutive integers for the two forms to differ over.
That gives $P (X < x) = P (X \leq x - 1)$ and $P (X > x) = P (X \geq x + 1)$ , so $P (X < 5) = P (X \leq 4)$ .
Complete the identity that turns an "at least" probability into a cumulative one:
$P (X \geq x) = 1 - P (X \leq______)$
The completed identity is:
$P (X \geq x) = 1 - P (X \leq x - 1)$
Subtracting the 1 is the step most often missed: $P (X \geq 10) = 1 - P (X \leq 9)$ and not $1 - P (X \leq 10)$ , because the value $X = 10$ has to stay inside the range.
How do you find $P (a \leq X \leq b)$ for a binomial variable from a cumulative function?
Subtract the cumulative probability below $a$ from the cumulative probability up to $b$ :
$P (a \leq X \leq b) = P (X \leq b) - P (X \leq a - 1)$
Using $a - 1$ rather than $a$ is what keeps the value $X = a$ inside the range, so $P (4 \leq X \leq 9) = P (X \leq 9) - P (X \leq 3)$ .
True or False?
For a binomial variable, $P (8 < X < 15)$ and $P (9 \leq X \leq 14)$ are the same.
True.
List the integers each one allows: strictly between 8 and 15 gives 9, 10, 11, 12, 13 and 14, and so does 9 to 14 inclusive.
Rewriting a range with $\leq$ at both ends is usually the first move, because a calculator's cumulative function works in those terms.
Which calculator function gives $P (X = x)$ for a binomial variable, which gives $P (X \leq x)$ , and what do you enter?
The binomial probability distribution function, often labelled Binomial PD, gives $P (X = x)$ , and the binomial cumulative distribution function, Binomial CD, gives $P (X \leq x)$ .
Both take the same three inputs: the value of $x$ , the number of trials $n$ , and the probability of success $p$ .
Some calculators ask for a lower and an upper bound instead, and then a lower bound of 0 gives the cumulative probability while a lower bound of $x$ with an upper bound of $n$ gives $P (X \geq x)$ directly.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Probability DistributionsNext:Normal Distribution