Level 3MathsAQAMathematical Studies (Core Maths)Paper 2A: Statistical TechniquesFlashcardsThe Normal DistributionThe Normal Distribution

The Normal Distribution (AQA Level 3 Mathematical Studies (Core Maths): Paper 2A: Statistical Techniques): Flashcards

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  • FrontThe Normal Distribution

    Define a continuous random variable.

Cards in this collection (22)

  • Define a continuous random variable.

    A continuous random variable can take any value within a range of infinitely many values.

    Continuous random variables usually measure something, such as height, weight or time.

  • True or False?

    For a normally distributed variable X, the probability that X takes one exact value is zero.

    True.

    Probability is the area under the curve, and a single value is a line of no width, so the area is zero.

    This is why \text{P} \left(X \le k\right) and \text{P} \left(X < k\right) mean the same thing for a normal distribution.

  • On a normal distribution curve, what does the area between x = a and x = b represent, and what is the total area under the whole curve?

    The area between them is the probability \text{P} \left(a < X < b\right).

    The total area under the curve is 1, because the variable is certain to take some value.

  • Complete the notation used when X follows a normal distribution:

    X \sim \text{N} \left(\_\_\_\_\_\_ , \_\_\_\_\_\_\right)

    The completed notation is:

    X \sim \text{N} \left(\mu , \sigma^{2}\right)

    The first number in the bracket is the mean and the second is the variance, so square root the second one whenever you need the standard deviation.

  • For a normal distribution, where is the curve's line of symmetry, and what does this tell you about the mean, median and mode?

    The curve is symmetrical about x = \mu, its mean.

    Because of that symmetry the mean, median and mode all take the same value, \mu.

  • A normal distribution curve has two points of inflection. Where are they?

    They lie one standard deviation either side of the mean, at x = \mu - \sigma and x = \mu + \sigma.

    They are the points where the curve stops bending one way and starts bending the other.

  • Fill in the missing percentages in these approximate results for a normal distribution:

    • About \_\_\_\_\_\_ of the data lies within one standard deviation of the mean

    • About \_\_\_\_\_\_ of the data lies within two standard deviations of the mean

    The completed results are:

    • About 68%, or roughly two-thirds, of the data lies within one standard deviation of the mean

    • About 95% of the data lies within two standard deviations of the mean

    Nearly all of the data, about 99.7%, lies within three standard deviations of the mean.

  • True or False?

    Changing the variance of a normal distribution moves the curve sideways without changing its shape.

    False.

    Changing the variance stretches the curve horizontally: a small variance gives a tall, narrow curve and a large variance a short, wide one.

    It is changing the mean that slides the curve sideways and leaves its shape alone.

  • What is the standard normal distribution, and what letter is used for it?

    The standard normal distribution is the normal distribution with mean 0 and standard deviation 1.

    It is written Z \sim \text{N} \left(0 , 1\right).

  • Complete the formula for standardising a value x taken from X \sim \text{N} \left(\mu , \sigma^{2}\right):

    z = \frac{x - \_\_\_\_\_\_}{\_\_\_\_\_\_}

    The completed formula is:

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

    You subtract the mean and then divide by the standard deviation, so square root the variance first if that is what the question gives you.

  • What does a z-value tell you about a data value, and what does a negative z-value mean?

    A z-value says how many standard deviations the data value lies away from the mean.

    A negative z-value means the data value is below the mean.

  • Define the notation \Phi \left(z\right).

    \Phi \left(z\right) is the probability that the standard normal variable Z is less than z, that is \Phi \left(z\right) = \text{P} \left(Z < z\right).

    It is an area, so it is always a number between 0 and 1, never a z-value itself.

  • The standard normal curve is symmetrical about z = 0. How does this relate \Phi \left(- z\right) to \Phi \left(z\right)?

    Symmetry gives \Phi \left(- z\right) = 1 - \Phi \left(z\right).

    The area to the left of - z equals the area to the right of z, and that is what is left when \Phi \left(z\right) is taken from the total area of 1.

  • The equation of a normal curve is far too complicated to integrate by hand. So how is the area under it actually found?

    The area has to be worked out numerically rather than by algebra.

    In practice that means either the cumulative normal distribution function on a calculator, or a table of the normal distribution function.

  • True or False?

    A "Normal Probability Density" function gives the probability that X lies below a given value.

    False.

    It gives the probability density at a single point, which is not a probability at all, and a single point has probability zero in any case.

    The function you want is the cumulative normal distribution, which returns the area between two bounds.

  • A cumulative normal function needs both a lower and an upper bound. How can you use it to find \text{P} \left(X > a\right), which has no upper bound?

    Use a as the lower bound and choose an upper bound far above the mean, since there is effectively no area left out there.

    More than four standard deviations above the mean is ample: beyond that the remaining probability is under 0.000032.

  • Complete the identity that turns a "between two values" probability into two one-sided ones:

    \text{P} \left(a < X < b\right) = \text{P} \left(X < \_\_\_\_\_\_\right) - \text{P} \left(X < \_\_\_\_\_\_\right)

    The completed identity is:

    \text{P} \left(a < X < b\right) = \text{P} \left(X < b\right) - \text{P} \left(X < a\right)

    Take the smaller area away from the larger one: doing it the other way round would give a negative probability.

  • To use a table of the normal distribution function, what must you do to a value x first, and why?

    Standardise it, turning x into a z-value.

    The table is built for the standard normal distribution Z \sim \text{N} \left(0 , 1\right) and knows nothing about your \mu and \sigma, whereas a calculator's cumulative function is told them directly and so needs no such step.

  • In an inverse normal calculation, what are you given and what are you finding?

    You are given a probability, together with the mean and the standard deviation.

    You are finding the value that has that probability lying below it, which is the reverse of working out \text{P} \left(X < a\right) from a value of a you already know.

  • True or False?

    If you know \text{P} \left(X > a\right) = p, you can put p straight into an inverse normal calculation that works from the left.

    False.

    A left-tail calculation expects the probability below a, so you must convert first, using \text{P} \left(X < a\right) = 1 - p.

    Some calculators offer a right-tail option instead, and with that selected p can be entered as it stands.

  • You solve \text{P} \left(X < a\right) = 0.27 for a. Without redoing any working, how can you tell whether your answer is sensible?

    Compare a with the mean.

    Since 0.27 is less than 0.5, under half the distribution lies below a, so a must come out smaller than the mean; an answer above the mean has to be wrong.

  • You read a value out of a table of percentage points of the normal distribution. What kind of quantity is it, and what must you do with it next?

    It is a z-value belonging to the standard normal distribution, not a value of X.

    Convert it back using the standardising formula rearranged as x = \mu + z \sigma.

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