1. S1: Chapter 9 The Normal Distribution Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 4 th March 2014

2. What does it look like? The following shows what the probability distribution might look like for a random variable X, if X is the height of a randomly chosen person. Height in cm (x) p(x) 180cm We expect this ‘bell-curve’ shape, where we’re most likely to pick someone with a height around the mean of 180cm, with the probability diminishing symmetrically either side of the mean. A variable with this kind of distribution is said to have a normal distribution.

3. What does it look like?

4. Height in cm (x) 180cm For a Normal Distribution to be used, the variable has to be: continuous With a discrete variable, all the probabilities had to add up to 1. For a continuous variable, similarly: the area under the probability graph has to be 1. To find P(170cm < X < 190cm), we could: find the area between these values. Would we ever want to find P(X = 200cm) say? Since height is continuous, the probability someone is ‘exactly’ 200cm is infinitesimally small. So not a ‘probability’ in the normal sense. Normal Distribution Q & A Q1 Q2 Q3 190cm170cm ? ? ? p(x) Q4 ?

5. Notation The random variable X......is distributed...

6. Z value The Z value is the number of standard deviations a value is above the mean. 100 IQ (x) Example IQZ 1000 1302 85 1654.333 62.5-2.5 ? ? ? ? ? p(x)  ?

7. You may be wondering why we have to look up the values in a table, rather than being able to calculate it directly. The reason is that calculating the area under the graph involves integrating (see C2), but the probability function for the normal distribution (which you won’t see here) cannot be integrated! A z-table allows us to find the probability that the outcome will be less than a particular z value. For IQ, P(Z < 2) would mean “the probability your IQ is less than 130”. (You can find these values at the back of your textbook, and in a formula booklet in the exam.) P(Z < 2) = 0.9772 ? Z table IQ (x) 100130 115 z=0 z=1z=2 Expand Minimise

8. Use of the z-table Suppose we’ve already worked out the z value, i.e. the number of standard deviations above or below the mean. Bro Tip: We can only use the z-table when: a)The z value is positive (i.e. we’re on the right half of the graph) b)We’re finding the probability to the left of this z value. 1 This is clear by symmetry. ? 2 ? 3 ? 4 ?

9. Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples). 1 2 3 4 5 6 7 8 9 10 ? ? ? ? ? ? ? ? ? ?

10. ‘Standardising’ Standardise

11. ‘Standardising’ Here’s how they’d expect you to lay out your working in an exam: The heights in a population are normally distributed with mean 160cm and standard deviation 10cm. Find the probability of a randomly chosen person having a height less than 180cm. No marks attached with this, but good practice! A statement of the problem. M1 for “attempt to standardise” Look up in z-table ? ? ? ?

12. Test your Understanding Steel rods are produced by with a mean weight of 10kg, and a standard deviation of 1.6kg. Determine the probability that a randomly chosen steel rod has a weight below 9kg. ?

13. Exam Questions Edexcel S1 May 2012 Edexcel S1 May 2013 (Retracted) Edexcel S1 Jan 2011 P(Z > -1.6) = P(Z < 1.6) = 0.9452 P(Z < -0.5) = P(Z > 0.5) = 1 – P(Z < 0.5) = 0.3085 P(Z > 1.6) = 1 – P(Z < 1.6) = 0.0548 ? ? ?

14. Quickfire Probabilities Example: P(X < 115) = P(Z < 1) = 0.8413 P(X < 107.5) = P(Z < 0.5) = 0.6915 P(X > 122.5) = P(Z > 1.5) = 1 – P(Z < 1.5) = 0.0668 P(X > 85) = P(X < 115) = P(Z < 1) = 0.8413 P(X < 106) = P(Z < 0.4) = 0.6554 P(X > 95) = P(X < 105) = P(Z < 0.33) = 0.6293 P(X 0.5) = 1 – P(Z < 0.5) = 0.3085 P(X 1.33) = 1 – P(Z < 1.33) = 0.0918 1 2 3 4 5 6 7 ? ? ? ? ? ? ? (For further practice outside class)

15. Probabilities for Ranges IQ (x) 100 z=0 96112 ? We can see that, in general: P(a < Z < b) = P(Z < b) – P(Z < a)

16. Quickfire Probabilities P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915 P(145 < X < 151) = P(3 < Z < 3.4) = P(Z < 3.4) – P(Z < 3) = 0.001 P(116 < X < 120) = P(1.07 < X < 1.33) = 0.9082 – 0.8577 = 0.0505 P(80 < X < 110) = P(-1.33 < X < 0.67) = 0.7486 – (1 – 0.9082) = 0.6568 P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486) – (1 – 0.9772) = 0.2286 1 2 3 4 5 ? ? ? ? ?

17. Summary so far Use Z-table must be <

18. The reverse: Finding the z-value for a probability ? ? ? ? ? ? ? ? ? ? ? ? ? For nice ‘round’ probabilities, we have to look in the second z-table. You’ll lose a mark otherwise. ?

19. The reverse: Finding the z-value for a probability What IQ corresponds to the top 22% of the population? State the problem in probabilistic terms. Find the closest probability in the z-table. ? ? 0.78 Bro Tip: Draw a diagram for these types of questions if it helps. z = 0.77 ‘Standardise’. ? ?

20. The reverse: Finding the z-value for a probability What IQ corresponds to the top 10% of the population? ‘Standardise’ Find the closest probability in the z-table. z = 1.2816 ? ? ? 0.9 z = 1.2816 State the problem in probabilistic terms. ?

21. The reverse: Finding the z-value for a probability What IQ corresponds to the bottom 30% of the population? We can’t look up probabilities less than 0.5 in the table, so manipulate: z = -0.5244 Convert back into an IQ. ? ? Find the closest probability in the z-table. Again, use the second table. ? ‘Standardise’ State the problem in probabilistic terms. ? ?

22. Test Your Understanding ? ? ?

23. Exercise 9C Page 184 1 6 7 9 ? ? ? ? ? ? ?

24. Quartiles IQ (x) P(X = x) Q 3 = 110 Q 1 = 90 ? ? Q 2 = 100 ? i.e. The upper quartile is two-thirds of a standard deviation above a mean (useful to remember!)

25. Edexcel S1 Jan 2011 Exam Questions Edexcel S1 May 2012 P(Z < z) = 0.6 z = 0.2533 (using 2 nd table) W = 162 + (7.5 x 0.2533) = 163.89975cm ? ? Edexcel S1 May 2010 a)P(D > 20) = P(Z > -1.25) = P(Z < 1.25) = 0.8944 b)P(Z z = 0.67 Q 3 = 30 + (0.67 x 8) = 35.36 c)Q1 = 30 – (0.67 x 8) = 24.64 ?

26. ? ?

27. ? First deal with P(X > 35) = 0.025Next deal with P(X < 15) = 0.1469 We now have two simultaneous equations. Solving gives: ? ?

28.   Test your understanding For the weights of a population of squirrels, which are normally distributed, Q 1 = 0.55kg and Q 3 = 0.73kg. Find the standard deviation of their weights. Due to symmetry,  = (0.55 + 0.73)/2 = 0.64kg If P(Z < z) = 0.75, then z = 0.67. 0.64 + 0.67  = 0.73  = 0.134   = 0.114  = 0.124  = 0.134  = 0.144 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.     = 2.77  = 2.78  = 79  = 2.80     = 76.15  = 76.25  = 76.35  = 76.45

29. Exam Questions Edexcel S1 May 2013 (Retracted) Edexcel S1 Jan 2011 Edexcel S1 Jan 2002 ? ? ?

30. Summary A z-value is: The number of standard deviations above the mean. A z-table is: A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1. P(IQ < 115) = 0.8413 We can treat quartiles and percentiles as probabilities. For IQ, what is the 85 th percentile? 100 + (1.04 x 15) = 115.6 We can form simultaneous equations to find the mean and standard deviation, given known values with their probabilities. B C D G H ? ? ? ? ? F P(a < X < b) = P(X < b) – P(X < a) ? A normal distribution is good for modelling data which: tails off symmetrically about some mean/ has a bell-curve like distribution. A ? E We need to use the second z-table whenever: we’re looking up the z value for certain ‘round’ probabilities. ?

31. Mixed Questions Edexcel S1 June 2001