Define continuous random variable .

A continuous random variable is a random variable that can take any value within a range , rather than only certain separate values. Continuous random variables usually measure something, so height, weight and time are all continuous. That is the contrast with a discrete random variable, which counts, and the difference decides which distributions can model the variable.

What has to be true of a real-life variable before a normal distribution is a sensible model for it?

It must be continuous , so it measures something, its distribution must be symmetrical and bell-shaped with a single mode , and the population needs to be large enough. So, for example , a variable produced by a random number generator cannot be modelled this way, because every value is equally likely and it has no mode. Nor can how long a human lives, because that distribution is not symmetrical.

True or False? A normal distribution cannot model height, because it allows any real value and a height cannot be negative.

False. It is true that a normal distribution is defined for every real number, but values more than about four standard deviations from the mean have a probability density of practically zero. So a normal model of human height puts a negligible probability on the impossible values and describes the realistic ones well, which is what makes it usable for quantities like height and weight that have a natural floor. A model does not have to be perfect to be useful, only good enough over the range that matters.

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Normal Distribution (Edexcel A Level Maths: Statistics): Flashcards

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Define continuous random variable.
A continuous random variable is a random variable that can take any value within a range, rather than only certain separate values.
Continuous random variables usually measure something, so height, weight and time are all continuous.
That is the contrast with a discrete random variable, which counts, and the difference decides which distributions can model the variable.
For a continuous distribution, what is $P (X = k)$ , and what follows from it?
$P (X = k) = 0$ for every value of $k$ , because probability is the area under the graph and a single value is a line with no width.
What has a probability is a range of values: the area between $x = a$ and $x = b$ is $P (a \leq X \leq b)$ , and the total area under the graph is 1.
What follows is convenient: since the endpoints contribute nothing, strict and weak inequalities give the same answer.
$P (X \leq k) = P (X < k)$
True or False?
In $X ~ N (μ, σ^{2})$ , the second number in the bracket is the standard deviation.
False.
The second number is the variance, $σ^{2}$ : the notation says so, but it is easy to read past.
So $W ~ N (50, 36)$ has variance 36 and standard deviation $\sqrt{36} = 6$ .
This matters because calculators ask for the standard deviation, so entering the second number straight from the bracket is a common error; square root it first, unless the variance is already written as a square, as in $N (20, 5^{2})$ .
A normal distribution is symmetrical about $x = μ$ . What two things follow from that?
The three averages all coincide:
$mean = median = mode = μ$
and half the area lies on each side of the mean:
$P (X < μ) = P (X > μ) = 0.5$
The second is worth having ready. It gives you a probability with no calculation, and it is often the quickest way to check that an answer is on the right side of the mean.
Complete the proportions of a normal distribution that lie within one, two and three standard deviations of the mean:
within $μ \pm σ$ , about $______$ % of the data
within $μ \pm 2 σ$ , about $______$ % of the data
within $μ \pm 3 σ$ , about $______$ % of the data
The completed proportions are:
within $μ \pm σ$ , about 68% of the data, which is roughly two thirds
within $μ \pm 2 σ$ , about 95% of the data
within $μ \pm 3 σ$ , about 99.7% of the data, which is nearly all of it
These are worth knowing by heart as a sense check. If a calculation says that 40% of the data lies within one standard deviation of the mean, something has gone wrong.
Where are the points of inflection on a normal distribution curve?
At $x = μ \pm σ$ , exactly one standard deviation either side of the mean.
Those are the two places where the curve stops bending one way and starts bending the other, as it changes from falling ever more steeply to falling ever less steeply.
This gives you a way to read $σ$ off a sketch: it is the horizontal distance from the mean to a point of inflection.
How does a normal curve change when $μ$ changes, and how when $σ^{2}$ changes?
Changing $μ$ translates the curve horizontally, moving the whole shape along without altering it.
Changing $σ^{2}$ stretches it horizontally: a small variance gives a tall curve with a narrow centre, and a large variance a short curve with a wide centre.
The reason a narrower curve has to be taller is that the total area is always 1, so squeezing the curve inwards must push it upwards.
What has to be true of a real-life variable before a normal distribution is a sensible model for it?
It must be continuous, so it measures something, its distribution must be symmetrical and bell-shaped with a single mode, and the population needs to be large enough.
So, for example, a variable produced by a random number generator cannot be modelled this way, because every value is equally likely and it has no mode.
Nor can how long a human lives, because that distribution is not symmetrical.
True or False?
A normal distribution cannot model height, because it allows any real value and a height cannot be negative.
False.
It is true that a normal distribution is defined for every real number, but values more than about four standard deviations from the mean have a probability density of practically zero.
So a normal model of human height puts a negligible probability on the impossible values and describes the realistic ones well, which is what makes it usable for quantities like height and weight that have a natural floor.
A model does not have to be perfect to be useful, only good enough over the range that matters.
How do you find $P (a < X < b)$ for a normal distribution?
The probability is the area under the normal curve between $x = a$ and $x = b$ , and since that curve is too complicated to integrate the area is found numerically, using the Normal Cumulative Distribution function on a calculator.
Often shortened to NCD, Normal CD or Normal Cdf, it needs four inputs: the lower bound $a$ , the upper bound $b$ , the mean, and the standard deviation.
Sketch the curve and shade the area you want before you start: it costs a few seconds and tells you whether the answer you get back is plausible.
True or False?
A calculator's Normal Probability Density function gives you $P (X = x)$ .
False.
It gives the probability density at that point, which is the height of the curve, not a probability.
For a normal distribution $P (X = x) = 0$ for every value of $x$ , so a function returning a non-zero answer cannot be giving you one.
The function you want is always the Normal Cumulative Distribution; the density function has no use in this course, so if your calculator offers you Normal PD or Normal Pdf, that is not it.
Your calculator wants both a lower and an upper bound, but you need $P (X > a)$ , which has no upper bound. What do you enter?
A value far enough above the mean that everything beyond it is negligible: more than four standard deviations above is accurate enough, and the easiest thing to type is a string of 9s, or $10^{99}$ .
It works because the probability of being more than three standard deviations above the mean is already less than 0.0015, and beyond four it is less than 0.000032, so the tail you are chopping off cannot affect the answer at the accuracy you are working to.
For $P (X < b)$ , do the same at the other end with a large negative lower bound.
A calculator will work out any normal probability directly, so when are results like $P (X > a) = 1 - P (X < a)$ still needed?
Whenever you cannot simply type the numbers in.
That happens when the mean or the standard deviation is unknown, when you have been given only a diagram rather than values, and when you are working with the inverse distribution and have a probability rather than a value.
The ones worth having ready are
$P (X > a) = 1 - P (X < a)$
$P (a < X < b) = P (X < b) - P (X < a)$
and they hold because the probability of a single value is zero, so nothing is lost or double counted at the boundaries.
You know that $P (X < a) = p$ and you need the value of $a$ . What do you use?
The Inverse Normal Distribution function, sometimes called InvN.
It runs the cumulative function backwards: the cumulative function takes a value and returns a probability, while the inverse takes a probability and returns a value.
Enter the area $p$ , the mean, and the standard deviation; if your calculator asks which tail, this is the left tail, because $p$ is the area to the left of $a$ .
For $W ~ N (50, 36)$ you are told $P (W > w) = 0.175$ and want $w$ . Complete the two things needed before the inverse normal function can be used:
the area to the left, $P (W < w) =______$
the standard deviation, $σ =______$
The completed values are:
the area to the left, $P (W < w) = 0.825$
the standard deviation, $σ = 6$
The area is converted with $P (W < w) = 1 - P (W > w)$ , because the function expects the area to the left; if your calculator offers a right tail, you can enter 0.175 directly instead.
The standard deviation is $\sqrt{36}$ , since the 36 in the bracket is the variance. Those two conversions give $w = 55.6$ to 3 significant figures.
What check should you always run on an answer from the inverse normal distribution?
Compare it with the mean, which tells you at once whether it is on the right side.
If $P (X < a)$ is less than 0.5 then $a$ must be below the mean, and if it is more than 0.5 then $a$ must be above it.
This catches the commonest mistake in inverse normal questions, which is working with the wrong tail; a quick sketch with the area shaded makes the comparison obvious.
What is the standard normal distribution, and how is it written?
The standard normal distribution is the normal distribution with mean 0 and standard deviation 1, and it is given its own letter, $Z$ :
$Z ~ N (0, 1^{2})$
Every normal distribution can be turned into this one by a horizontal translation and a horizontal stretch, which is why a single standard distribution is enough to describe them all.
Some writers use $Φ (z)$ to mean $P (Z < z)$ .
Complete the formula that standardises a value of $X ~ N (μ, σ^{2})$ :
$Z = \frac{X -______}{______}$
The completed formula is:
$Z = \frac{X - μ}{σ}$
Each operation does one of the two transformations: subtracting the mean slides the distribution along so that it is centred on 0, and dividing by the standard deviation stretches it so that its standard deviation becomes 1.
So a $z$ -value measures how many standard deviations a value is away from the mean, and a value below the mean gives a negative $z$ .
The rearranged form $x = μ + σ z$ is often the more useful one.
True or False?
Standardising a normal distribution changes the probabilities.
False.
Standardising rescales the horizontal axis, and the probabilities go with it unchanged:
$P (X < a) = P (Z < \frac{a - μ}{σ})$
The area under the curve is what carries the probability, and translating and stretching the curve rearranges that area without creating or destroying any of it.
This is what makes the standard normal distribution useful rather than merely tidy: a question about any normal distribution can be turned into one about $Z$ .
A calculator handles any normal distribution directly, so what is the standard normal distribution actually needed for?
For questions where the mean or the standard deviation is unknown: a calculator cannot be given an unknown parameter, so those questions cannot be done by entering numbers.
What you can do is work in terms of $Z$ , whose mean and standard deviation are always 0 and 1, find the $z$ -value that goes with the given probability, and then use
$z = \frac{x - μ}{σ}$
as an equation in the unknown.
The same idea makes the standard normal distribution the natural setting for a hypothesis test on the mean of a normal distribution.
Times are modelled by $N (μ, 4^{2})$ and 10% of them are less than 12 minutes. How do you find $μ$ ?
Turn the probability into a $z$ -value, then solve for $μ$ .
Since $P (T < 12) = 0.1$ , the inverse normal on the standard distribution gives $z = - 1.281551 \dots$ , and substituting into $z = \frac{x - μ}{σ}$ :
$- 1.281551 \dots = \frac{12 - μ}{4}$
$μ = 12 + 4 (1.281551 \dots) = 17.1 minutes$
The negative $z$ is what makes the mean come out above 12, which is right: if only 10% of times are below 12 minutes, the average time must be longer than that.
Both $μ$ and $σ$ are unknown. What will the question give you, and how do you use it?
It will give you two probabilities, for two different values of $x$ , because two unknowns need two pieces of information.
Each one gives a $z$ -value in the usual way, so you get two equations, and writing them in the form
$x = μ + σ z$
makes them a straightforward pair of simultaneous equations in $μ$ and $σ$ , which a calculator can solve.
The thing to be careful about is keeping each $z$ -value with its own value of $x$ : swapping them gives a pair of equations that looks perfectly reasonable and is wrong.
How do you know whether a $z$ -value should be negative, and how many decimal places should you keep?
Sketch the curve first, marking the mean and the known value: if $x$ lies below the mean then its $z$ -value is negative, and getting that sign wrong is the commonest way these questions go astray.
Make sure the inequality for $Z$ points the same way as the one for $X$ : if you are told $P (X > x)$ , you want $P (Z > z)$ .
Keep plenty of decimal places in the $z$ -value, at least one more than the accuracy you want in the final answer, since the $z$ gets multiplied by $σ$ and rounding it early is magnified.

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