€¦ · Web view"In terms of wizarding prowess, you're three standard deviations above the mean, Harry."The Harry Potter Guide To S1. Last updated - 8th April 2014

The Harry Potter Guide To S1Last updated - 8th April 2014

Chapter 0 - Guide To Using Your Casio Calculator _

Practice being able to calculate these with your calculator - you should not not rely on them to calculate your answer in the first place, but allows you to check if you've got your answer correct. We'll use the following data sets:

Lengths of wands: 12.3cm, 15.7cm, 20.4cm, 21.3cm, 29.2cm

Weights of owls w (kg) 3≤w<5 5≤w<6 6≤w<10 10≤w<15Frequency 5 8 2 1

Hours Revised 7 10 17 8Potions test score 75 86 92 76

Harry Tip #1: Be careful to note how many variables there are in the problem.

The first data set is obviously just one variable (time). The second table (where the data is grouped) is still one variable, but has frequency information. On your calculator, press MODE then choose 2 for STAT, and finally 1-VAR for "1 variable". The get the frequency column, go to SETUP (SHIFT -> MODE), press the down key to get to next page on the menu, choose 3 for STAT, then finally select 1 for 'FREQUENCY ON'. You may as well leave this mode on, since the frequency value defaults to 1 when you don't specify it (i.e. you just have one instance of each listed piece of data).

For the second data set, you must not choose the 'A + BX' mode and use your second Y column for the frequencies, as you can't treat the frequency data as just a second variable.

For the last data set, since there's two variables, from SETUP -> STAT, choose 2 for 'A + BX' for linear regression with two variables. This is the equation of the straight line, and this is because for the purposes of S1, your variables are assumed to have a linear relationship (i.e. roughly follow a line of best fit) when calculating both your Product Moment Correlation Coefficient (which measures how well your data fits a straight line) and your regression line. The other modes, which you won't use at A Level, allows your variables to have different relationships, e.g. for population growth (which grows/falls exponentially), the 'A.B^X' mode would be more appropriate as y=a ⋅bx is an exponential function relating the two variables x and y.

Entering your data

Enter each value for your first variable, pressing = each time to get to the next row. If you have a second variable or frequencies to enter, use the arrow keys to navigate back to the top right of your table. Once done, press the AC key to 'bank' your table. It's now stored in memory.

Calculating a statistic

Press SHIFT -> 1 for 'STATISTIC'. Choose 'VAR' for calculate either the mean or variable of either of your variables (x and where relevant y). This will then use this quantity in your current calculation, which you can further manipulate if you like. Then press = to get the value.

Here's a summary of what you can calculate:

www.drfrostmaths.com

"In terms of wizarding prowess, you're three standard deviations above the mean, Harry."

Set 1

Set 2

Set 3

VAR Standard deviation σx (don't use sx), number of data values n, and the mean x , y .

SUM The sum of the values of your variable Σ x , Σ y or the sum of the squares Σ x2, or the sum of the products Σ xy . The second and third of these are obviously crucial when you're calculating Sx x, Syy and Sxy, which your calculator is unable to do directly.

Important note: Note that when you have a frequency row/column, Σ x is actually calculating Σ fx, and Σ x2 is actually calculating Σ f x2 because the values are being duplicated so each copy of the value is effectively being treated as a separate value of x. We'll discuss this later.

REG Only available when you're in 2-variable mode. Allows you to calculate the y-intercept and gradient of your line of best fit, a and b in y=a+bx and the Product Moment Correlation Coefficient r .

Now have a go at calculating the following for the 3 data sets above:

Data Set 1 The mean, variance and standard deviation of the length of wands.

Answers: σ=5.73 cm, σ 2=32.81, x=19.78cm

Data Set 2 An estimate of the mean and standard deviation of the weight (recall that you need to use the midpoints of the class intervals).

Try calculating the standard deviation by typing

in the full formula for variance, σ 2=Σ x2

n−x2 by

repeated use of the 'STATISTIC' button before pressing =. Verify that this gives you the same value as when you use your calculator to calculate σ 2 directly.

w=5.78 kg, σ w=2.11kg

Data Set 3 Try and find Σx2, Σ y2, Σxy , Σ x and Σ y. Hence findSxx, Syy and Sxy using these values. Find r directly and the linear regression line.

Σx=42, Σ x2=502, Σ y=392, Σ y2=27261, Σ xy=3557

r=0.926

y=64.607+1.680x


5kg 6kg5

1311.52

?

Chapters 2/3 - Data: Location and Spread _

Hagrid Tip #1: When you have a discrete list of items, to find the median/quartiles, find 14 ,

12 , 34 of the number of items n, and then round up and use that numbered item. The one

exception is when you have a whole number after dividing, in which case use this item and the one after.

Example: 2, 4, 6, 8, 10, 12There are 6 items. For the median 6/2 = 3. This is a whole number, so use 3rd and 4th item (midpoint is 7). For the LQ, 6/4 = 1.5. This rounds up to 2, so use the second item (4).

Hagrid Tip #2: When you have grouped continuous data, and you're finding the quartiles/percentiles, DO NOT ROUND to find the item number - just keep it as it is. Use linear interpolation to find an estimate for your quartile/percentile.

Example: Consider our third data set again. We add a cumulative frequency row:

Weights of owls w (kg) 3≤w<5 5≤w<6 6≤w<10 10≤w<15Frequency 5 8 2 1Cumulative Frequency 5 13 15 16

To calculate the Lower Quartile: n=16, so 16 /4=¿ 4th item. The 4th item occurs within the first 5 items. We can put the cumulative frequency at the start and end of the matching

interval, as well as the item we're interested in. We can also put the class boundaries on the bottom side of the line.

We're clearly 4/5 of the way along the line, so we go 4/5 of the way along from 3kg to 5kg:

Q1=3+( 45 ×2)=4.6kgTo calculate the 72th Percentile, P72:

72% of 16 = 16×0.72=¿ 11.52th item. This doesn't occur within the first 5 items but does occur within the first 13 items, so we know P72 is in the 5≤w<6 weight interval.

Thus P72=5+(6.528 ×1)=5.815 kg.

Hagrid Tip #3: Be vigilant of gaps in class intervals vs no gaps, and of dark wizards.Suppose we instead had the following data:

Weights of owls w (kg) 3−5 6−8 9−13Frequency 3 4 5Cumulative Frequency 3 7 12


3kg 5kg

0 54

?

5.5kg 8.5kg

3 76

?

We now have gaps! If you don't adjust the class intervals accordingly (i.e. the class widths would be 3, 3 and 5), you'll get absolutely no marks for your linear interpolation. On the plus side

however, Voldemort would consider enlisting you as a Death Eater.For example, to find the median:

n=12 so use 12÷2=¿ 6th item.

Q2=5.5+( 34 ×3)=7.75 kg Hagrid Tip #4: Try and memorise the mnemonic for the formula for variance, and how the

formula results from it, rather than memorise the formula itself. Hagrid Tip #5: Make sure you understand the difference between Σ x2 and (Σ x )2. Hagrid Tip #6: Use your calculator to check your value of the variance (see Calculator Tips

above).

The mnemonic for variance: "The mean of the squares minus the square of the mean" ("msmsm"). This gives:

1. Ungrouped data: σ 2=Σ x2

n−(Σ xn )

2

∨σ2=Σ x2

n−x2 (since x= Σ x

n )

2. Grouped data: σ 2=Σ fx 2

Σ f−( Σ fxΣ f )

2

. Don't get these two mixed up!

You should not however think of these formulae as different. Σ f clearly means the same as n. And Σ fx still means the (estimated sum) of the values (using the midpoints x of the class intervals). Confusingly, when exam questions use a variable for the values, say w , but the data is grouped, Σw still refers to the total of all the values with the frequencies factored in. This is likely to be different to Σ fx , because the latter is an estimate of the total using the midpoints of the grouped data, whereas Σw is the exact total of the values before the data was grouped and information was lost (see Edexcel Jan 2011 Q5 for example). This just means that if you wanted the mean of w and you

were given Σw, then w= Σwn , and ignore the grouped frequency table you were given.

Hagrid Tip #7: Don't forget to square root when you're finding the standard deviation from the variance!

Hagrid Tip #8: Check that your standard deviation looks sensible. Standard deviation roughly means "the average distance from the mean". So if your standard is 10 times too large say, then you know you've gone wrong.

Coding:

1. However you code your variable (adding, dividing, etc.) you do the same to the mean.2. Adding or subtracting to your variable doesn't affect the spread (variance/standard

deviation). This intuitively makes sense: were everyone to get exactly 50cm taller by standing on a chair, the heights are just as spread out.


Area Frequency

3. Multiplying or dividing affects the standard deviation in the same way. If you double the heights, you double the standard deviation. You halve the heights, you halve the standard deviation.

4. For variance though, we have to square the factor difference. For example, if the values tripled and hence σ becomes 3σ , then the variance is (3σ )2=9σ , i.e. the variance becomes 9 times larger.

Hagrid Tip #9: Make sure you check whether you're finding the new mean/variance/standard deviation after coding, or the original mean/variance/standard deviation before coding.

Chapter 4 - Data: Location and Spread _

Box Plots

Remember that you need to calculate your outlier boundaries, which are generally 1.5 Interquartile Ranges above the UQ or below the LQ. You will always be told in the exam question however how the outlier boundary is defined.

Buckbeak Tip #1: There's two possibilities for the end points of the whiskers when there's an outlier on that end, and mark schemes accept both: either use the outlier boundary itself, or the smallest/greatest value which is not an outlier.

Buckbeak Tip #2: You must explicitly show your calculation for the outlier boundaries. There's marks specifically for this in the mark scheme, and if you display your whiskers slightly incorrectly, you'll risk losing all marks.

Stem and Leaf Diagrams

You may be asked to calculate the interquartile range. In which case, just remember that you have a discrete list of items, and hence choose the items to use for the quartiles in the correct way.

When asked to compare the two sets of data in a back-to-back stem and leaf diagram, they're expecting things like "the boy's scores tend to be higher than the girls".

Histogram

Pretty much all histogram questions boil down to this simple diagram:

i.e. You're identifying the scaling (k ¿ from area to frequency. At GCSE you could always assume that k=1, i.e. area is EQUAL to frequency. Identifying k may come from either using the total area and total frequency given (when frequencies of individual intervals are not available) or from the known frequency and area of a particular bar. Once k is known, you can use it to calculate frequencies for any area.Use my slides for practice: http://www.drfrostmaths.com/resource.php?id=11371

Buckbeak Tip #3: If you're not given a frequency density scale on the histogram on the y-axis, and only know the total frequency, just add any frequency density scale. If you know


the frequency of a particular bar, it's generally easiest to set the scale such that

frequency density= frequencyclasswidth .

Buckbeak Tip #4: As before, you must check if the class intervals have gaps! If so, ensure you use the correct class intervals when calculating frequencies/frequency densities.

Buckbeak Tip #5: When asked to find the mean, median, quartiles or variance of a histogram, first use the histogram to generate a grouped frequency table. Then use this table as you usually would to calculate these statistics.

Buckbeak Tip #6: When asked why a histogram is an appropriate means of displaying the data, the words they're looking for are 'continuous data/variable', and nothing else.

Skew

Remember that there's 3 ways in which you calculate skew:

1. For histograms or probability distributions, just observe the shape. If the 'tail' is in the positive direction, you have positive skew. If it's in the negative direction, you have negative skew.

2. Use the quartiles. If the right box of the (implied) box plot is wider, i.e. Q3−Q2>Q2−Q1, then you have positive skew. If the left box is wider, you have negative skew. If they're the same width, you have no skew.

3. Use the mean and median. The way to remember which order means which type of skew is to think of salaries: Large salaries in the positive tail drag up the mean but not the median, hence mean>median means we have positive skew.

On the rare occasion, you have both the quartiles and mean available. In which case, choose either (2) or (3) to find what type of skew you have. Otherwise, the choice should be clear based on the data available.

Buckbeak Tip #6: For 2 mark questions which ask you to comment on skew, you get one mark for saying 'negative/positive/no skew', and the other mark for given a valid reason (e.g. mean>median).

Chapter 5 - Probability _

If A and B are independent, then A does not affect B and vice-versa. If A and B are mutually exclusive, then A and B can’t happen at the same time. These are completely separate things – one is not the opposite of the other!

Laws

If A and B are independent:

P (A|B )=P(A)(as the probability of A is not affected by B)

P (A ∩B )=P (A )P(B)(If you're asked to show that two events are independent, then show this equality holds)

If A and B are mutually exclusive:

P (A ∩B )=0 P (A∪B )=P ( A )+P (B)

In general:


P (A|B )= P (A∩B )P (B )

(I remember this by ‘the intersection divided by the thing I’m conditioning on’)(Notice that if A and B are independent, then

the RHS simplifies to P(A))

Addition Rule: P (A∪B )=P ( A )+P (B )−P ( A∩B )

(Remember this by thinking about two overlapping circles – we need to subtract the overlap)(Notice that if A and B are mutually exclusive,

P (A ∩B )=0, so we get our earlier formula)(If A and B are independent, this reduces to

P (A∪B )=P ( A )+P (B )−P ( A )P (B )


McGonagall Tip #1: Mutually exclusive events are indicated by separated non-overlapping circles in a Venn Diagram. Independence does not affect the Venn Diagram.

McGonagall Tip #2: If you’re not told two events are mutual exclusive, then for the purposes of the Venn Diagram, you have to assume that they are not mutually exclusive, i.e. they overlap.

McGonagall Tip #3: You can often determine probabilities by constructing a Venn Diagram and filling in the missing probability in regions by simple adding/subtracting. Other times, this approach doesn’t work.

McGonagall Tip #4: You can treat probabilities algebraically. e.g. P (A )−0.7 P ( A )=0.3P ( A )

McGonagall Tip #5: If you’re told that A and B are independent, immediately write out that P (A ∩B )=P (A )P (B ) using whatever probabilities you’re given. Same for mutual exclusivity. It’ll help you visualise the probabilities you have available to determine others you don’t know.

McGonagall Tip #6: Do you have a mixture of P(A∩B), P (A∪B ) and P(A) (or P (B ))? You should write out the Addition Rule and see if it helps. McGonagall Tip #7: Note that given some event, the probabilities add up to 1. So:

P (A|B )+P (A '|B )=1 and P (A|B' )+P ( A '|B' )=1Some people incorrectly assume:

P (A|B )+P (A '|B' )=1 McGonagall Tip #8: As per GCSE, a suitable tree diagram can work wonders. Remember that

the second level of branching and onwards are conditional probabilities.

Dealing with more complicated tree questions

Suppose you have a tree like the one below:


Then how would we calculate the following?

P (C ) As per GCSE, we just find all the paths which match this description (i.e. where C is true) and add the probabilities of each path:

P (C )=P ( A∩B∩C )+P ( A∩B'∩C )+P ( A'∩B∩C )+P(A'∩B '∩C)

P(A∩C ') P (A ∩C' )=P ( A∩B∩C ')+P (A∩B'∩C ' )

P (B ) Note that we need not even consider the event C because it occurs after B:

P (B )=P (A∩B )+P(A '∩B)

P (C|B ) Seeing the conditional probability, you should immediately go for your formula for conditional probability!

P (C|B )= P (B∩C )P (B )

=P ( A∩B∩C )+P ( A'∩B∩C )

P (A∩B )+P ( A'∩B )

Chapter 6 - Correlation _

Recall that r=1 is 'perfect positive correlation', r=0 is 'no correlation' and r=−1 is 'perfect negative correlation'. Anything below -0.7 or above 0.7 is considered to be strong correlation.I'm going to presume here you can plug values into your Sxx, Syy, Sxy and r formulae. But things I see go wrong:

Lupin Tip #1: In Sxx=Σ x2− (Σ x )2

n, I've seen people forget to square the Σ x , or mix the

formula up with the one for variable, and do Sxx=Σ x2−( Σ xn )2

. This formula is clearly

wrong because the n in the denominator gets squared when it shouldn't. Lupin Tip #2: You can use your calculator to directly calculator r if you're given the original

data (see the beginning of this guide). Make sure however you still show your calculations for Sxx and so on for the purposes of evidencing working. However, generally you're generally provided with certain sums in the exam to save you time, so you may not be able to enter the original data directly.

Here are some potential 'explaining' questions you might encounter:

Lupin Tip #3: If you're asked whether your correlation coefficient supports some assertion, just comment on whether your value is close to -1, 0 or 1. If someone is claiming that house prices falls with distance from central London, then their assertion is justified if you have a correlation coefficient close to -1 (i.e. negative correlation).

Lupin Tip #4: If you're asked to give an 'interpretation' of your correlation coefficient, this doesn't mean to say whether it's negative or positive, but to say what it actually means in


words. e.g. "Higher towns have lower temperature/temperature decreases as height increases". i.e. You're asked to state to what happens to one variable as the other increases.

Coding

r Completely unaffected by any multiplications, divisions, additions or subtractions in the coding.

Sxx S yy Since Sxx=nσ2 , Sxx is affected by coding in the

same way as variance. So if the value is doubled in coding, Sxx becomes 22 times bigger. The same applies to Syy.

Sxy As above, addition and subtraction in the coding has no effect. If x is scaled by a factor of k and y by a factor of q, then Sxy is scaled by a factor of kq. e.g. If Sxy=10 and a=4 x+1 and b=3 x−2, then Sab=Sxy×4×3=120.

Chapter 7 - Regression _

Regression in general means to find the parameters of a model which best explains the data. In the case of linear regression, the model is a straight line, and the parameters are its y-intercept and gradient, which are set to as to minimises the total (squared) error between the predicted y-value on the line and y-value on each data value.The formulae for the equation of the least squares regression line are given in the formula booklet, but they're worth memorises:

y=a+bx. Notice that unlike y=mx+c, we put the y-intercept term first on the RHS. This is so that when we extend say to quadratic regression, i.e. fitting a line y=a+bx+c x2, a still means the y-intercept.

b=SxySxx

. I remember this by the fact that 'xy' are the sex chromosomes for a man, and 'xx' for

a woman, and the man comes first (a little bit misogynistic I know, but easy to remember!) a= y−bx . This is easily remember by just rearranging y=a+bx above to make a the

subject, and then replacing x with x and y with y . This suggests that the point ( x , y ) is on your regression line. Remember that x just means the mean of x and is calculated using

x= Σ xn .

Snape Tip #1: If asked to interpret your gradient, say something like "As [my x variable] increasing by 1, the [your y variable] increases/decreases by [the gradient]", e.g. "as the height increases by 1m, the temperature decreases by 3 degrees".


Snape Tip #2: It's good to be clear from the outset what's your explanatory variable and what's your independent variable. This is crucial when you calculator your gradient, since if you get your

variables mucked up, you might do b=S xy

S yy by mistake. These will be rarely labelled using the

variable letters x and y, so ensure you work out which is which. If for example your had the water depth d and the pressure p, clearly p depends on d , so d is your x (explanatory) variable and p your y (dependent) variable. Don't write Sxy (since x and y don't exist in this context), use Sdp instead.Snape Tip #3: Remember that for coding, your just replace your variables using the expressions given. e.g. If you have the line y=3+2x and you used the coding x=3a+4, y=b−1, then just substitute: b−1=3+2 (3a+4 ) and simplify. Piece of cake.Chapter 6-7 - Interpretation Questions for Correlation/Regression _

Interpretation questions are a bit of a pain if you don't know the examiner is looking for, and often worth a lot of marks (anything up to 4!). Here's everything you need to know:

Question Answer format Comments

Interpret r "The graph show a ___ correlation.As ____ increases, ____ increases/decreases."

The key here is saying how the dependent value changes as the explanatory variable increases. The question asks you to 'interpret', so it's key here you describe what it explains using NON-STATISTICAL LANGUAGE. Saying "Positive correlation" alone would be DESCRIBING the correlation.

Explain why this diagram would support the fitting of a regression line of y onto x

"The variables have a linear relationship, i.e. the points are close to the implied straight line of best fit."

Interpret the gradient/slope of the line/interpret b

As (x) increases by 1, (y) increases/decreases by ___.

Again, the key word here is 'interpret'. Example "As the height increases by 1m, the temperature decreases by 0.1 °C

Interpret the y-intercept/interpret a

The value (y) takes when (x) is 0.

e.g. "A score of 55 would be obtained if no hours of revision are done".

Extrapolation "Reliable [1 mark] because the value is within the range of the data [1 mark].""Reliable because the value is close to the range of the data".

The 'tricksey' case here is when the value is just outside the data range. The examiner is expecting you to state the your regression equation is still


"Unreliable because the value is outside the range of the data, i.e. we are extrapolating."

reliable.

Which is the explanatory variable? Explain your answer.

"(x) is the explanatory variable because (x) influences (y)".

Explain method of least squares

"We minimise the square of the residuals" (draw a diagram)

You may wish to go back to your textbook to understand what's going on here. But the principle is that we're finding the error between each data value and the predicted value (based on the model), squaring the error (to ensure the error is positive) then adding them up. The gradient and y-intercept of the straight line is set to minimise this total error.

Chapter 8 - Discrete Random Variables _

A discrete random variable has two ingredients:

1. A set of outcomes.2. A probability distribution mapping outcomes to probability. This has two flavours:

a. A 'probability distribution' is represented either as a table or as a graph.b. A 'probability function' allows you to have a more complex expression that

calculates the probability, as opposed to a probability distribution where each outcome and its probability are explicitly stated. These might be specified as a 'piecewise function', where different functions are used for different ranges of

outcomes, e.g:

p ( x )={k ( x+1 ) x=1,2,3,40 otherwise

P (X=x ) is the longhand for writing a probability. It means "the probability that a random variable X has the value x. p(x ) is the shorthand. Note the use of lowercase p.A cumulative distribution function meanwhile gives the probability up to a certain value, i.e. F ( x )=P (X ≤ x ).

Finding the value of a missing variable using a probability function or cumulative distribution function

p ( x )={k ( x+1 ) x=1,2,3,40 otherwise

Probabilities must add up to 1. So just plug in each value of x into your probability function:

2k+3k+4k+5 k=1


So k= 114

F ( x )={k ( x+1 ) x=1,2,3,40 otherwise

It's clear F ( 4 )=1, because it's certain that we'll have an outcome up to 4.

This gives us 5k=1, so k=15

Finding a probability distribution from a cumulative distribution functionSince the cumulative distribution function is the running total of the probability, finding the difference tells us what we added to the running total each time.Expected ValueExpected value just means the 'mean' of the random variable. So for a fair die, E (X )=3.5 because we expect to see an outcome of 3.5 on average. For this reason E (X ) is sometimes represented as μ. We just sum the product of each outcome with its probability:

E (X )=Σ x p (x)Finding the value of missing variables using a probability distribution and a given expected valueWe form two simultaneous equations by:

1. Noting that the probabilities of a probability distribution add up to 1.2. Using the provided expected value.

e.g. Given that E (X )=3 and that X has the probability distribution below, determine a and b.

x 1 2 3 4

P (X=x ) a a b 0.1

2a+b+0.1=11a+2a+3b+0.4=3Then just simplify and solve.VarianceThe mnemonic "the mean of the square minus the square of the mean" still applies!

Var (X )=E (X2 )−E ( X )2

If you can't distinguish between E (X2 ) and E (X )2, then it's likely you're one of those people who

mix up Σx2 and (Σ x )2. E (X2 ) means you find the expected value as before, except you replace each

outcome with itself squared - the probability is always unaffected, i.e. E (X2 )=Σ x2 p ( x ).CodingThe same rules apply as before. For expected value, whatever we did to the random variable, we do to the expected value. Adding/subtracting has no effect on variance, and when we multiply or divide, we square this value.Examples: E [1−3 X ]=1−3 E[X ] Var [1−3 X ]=9Var [X ]Chapter 9 - Normal Distribution _


A normal distribution is perfect for data which is symmetrically distributed about some mean and the probability tails off as we get further from the mean.

Some preamblez-values and z-tablesThe z value means the number of standard deviations above the mean. For IQ for example, where

X N (100,152 ), and IQ of 130 corresponds with a z-value of 2, and an IQ of 85 with a z-value of -1.

z= x−μσ

This makes sense: x−μ is how far about the mean a value is, so dividing it by σ tells us how many standard deviations above the mean we are.If X is the original random variable, converting it to a new random variable Z using the formula above is known as standardisation. Z is a normally distributed variable where Z N (0,12 ), known as the standard normal distribution. The reason μ=0 and σ=1 is because then if the z value is say 3, then this IS the number of standard deviations above the mean, because 3 is 3 lots of 1 above 0. Thus in converting X to Z, each value of the random variable now represents the number of standard deviations above the mean. So an IQ of 130 gets converted to 2, and so on.A z-table allows us to find out the probability of being up to a certain z value, i.e. P (Z<z ), where z is a specific value of Z. Looking up z=3 in the table would determine the probability of being up to 3 standard deviations above the mean; in the case of IQ, it would tell us the probability of having an IQ less than 145.

Manipulating z-probabilitiesJust think about the graph of a normal distribution when manipulating probabilities. P (Z←2 ) represents the probability of being less than two standard deviations below the mean, e.g. having an IQ less than 70. Due to symmetry, this is the same probability as being at least 2 standard deviations above the mean, e.g. having an IQ greater than 130. i.e. P (Z←2 )=P (Z>2 ).Similarly, since the probability of having an IQ above 130 is one minus probability of having an IQ below 130, we have that P (Z>2 )=1−P (Z<2 ).In order to be able to use the z-table, we (a) need our z-value to be positive and (b) the inequality to be <. Examples: P (Z>−3 )=P(Z<3). P (Z>4 )=1−P (Z<4 )P (Z←0.5 )=P (Z>0.5 )=1−P (Z<0.5 ). Practice this manipulation!

Finding a probability"Let X represent the IQ of a randomly chosen person, where X N (100,152 ). Find the probability that a randomly chosen person has an IQ below 96".

Step 1: Express your information as a probability. P(X<96)

Step 2: Standardise, i.e. convert your X value to a Z value, using the formula (or common sense). Ensure you get the sign of z correct.

P(Z< 96−10015 )=P(Z←0.27)

Step 3: Manipulate your z probability so that the ¿ P (Z>0.27 )=1−P (Z<0.27 )


z value is positive and you're using <.

Step 4: Look up in z table. ¿1−0.6026=0.3974

Step 5: Check that your answer looks sensible. (Yes, the answer looks sensible because we expected slightly less than half the population)

Finding a value corresponding to a probabilityExample 1: "Find the lower quartile for IQ, given that X N (100,152 ) again."

Step 1: As before, express your information as a probability.

P (X<Q1 )=0.25

Step 2: Normalise. We don't like the probability of 0.25 because it won't be in our z-table (it's less than 0.5). A quick fix is to make the z-value minus, which allows us to do 1 minus the probability (as per on the right).

P (Z<z )=0.25P (Z← z )=0.75

Step 3: Ordinarily, we'd look up the closest z-value that gives that probability, which turns out to be 0.67. For certain 'nice' probabilities (e.g. 2.5%, 5%, 10%), we MUST look in the second z-table, otherwise you lose marks. However, in this case, 0.75 is not there, so use first table.

−z=0.67 z=−0.67This makes sense as we expect our z value to be negative as it's below the mean.

Step 4: Convert back into an X value by using the formula for z:

Q1−10015

=−0.67→Q1=89.95

I personalise find it easier to write Q1=100−(0.67×15 ) as it expresses the fact that Q1 is 0.67 standard deviations below the mean (i.e. 0.67 lots of 15 below 100). However, since the form at the top of this box is what's listed for the method mark, it's probably safer to do it that way.

Example 2: "Find the IQ for which 10% of the population exceeds, given that X N (100,152 ) again."

Step 1: As before, express your information as a probability. Clearly if 10% of people have an IQ higher than this value, then there's 90% below (recall that we always want the probability of being below some value on the right half of the graph in order to use the z-table).

P (X<x )=0.9

Step 2: Standardise. No manipulation of our z probability required as we have < and a probability greater than 0.5.

P (Z<z )=0.9


Step 3: This time we can use the second z-table because 0.1 is on (note that confusing, the second table gives us the probability of being ABOVE some z-value, i.e. P (Z>z )).

z=1.2816

Step 4: As before, use the formula for z to convert back into an IQ. 1.2816= x−100

15→x=119.224

Finding the missing mean or standard deviationYou're highly advised to do lots of practice questions on this until it's second nature!Example: "Only 10% of maths teachers live more than 80 years. Triple that number live less than 75 years. Given that life expectancy of maths teachers is normally distributed, calculate the standard deviation and mean life expectancy."

Step 1: As before, express your information in terms of probabilistic statements. Notice we've turned the "10% above" into "90% below".

P (X<80 )=0.9P (X<75 )=0.3

Step 2: Deal with first. P (X<80 )=0.9P (Z<z )=0.9z=1.28161.2816=80−μσ

Step 3: Deal with second. P (X<75 )=0.3P (Z<z )=0.3P (Z← z )=0.7−z=0.5244z=−0.5244

−0.5244=75−μσ

Step 4: We have two simultaneous equations, so solve!

80−μ=1.2816σ75−μ=−0.5244 σSubtracting the two equations:

5=1.806σ→σ=2.769μ=76.451


I find your S1 efforts most... a--sing.

Harry Potter stats puns (c) Lord Voldermort.Copyright infringement punishable by Cruciatus.


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€¦ · Web view"In terms of wizarding prowess, you're three standard deviations above the mean, Harry."The Harry Potter Guide To S1. Last updated - 8th April 2014