1. DP Studies Y2 Chapter 10: Normal Distribution

2. Contents: A. The normal distribution B. Probabilities using a calculator C. Quantiles or k-values

3. Opening Problem

4. A continuous random variable is a variable which can take any real value within a certain range. We usually denote random variables by a capital letter such as X. Individual measurements of this variable are denoted by the corresponding lower case letter x.

5. For example, the probability that an egg will weigh exactly 72.9 g is zero. If you were to weigh an egg on scales that weigh to the nearest 0.1 g, a reading of 72.9 g means the weight lies somewhere between 72.85 g and 72.95 g. No matter how accurate your scales are, you can only ever know the weight of an egg within a range. ( think of a number line where you have all real numbers ) So, for a continuous variable we can only talk about the probability that an event lies in an interval, and: P(a < X < b) = P(a < X < b) = P(a < X <b) = P(a < X < b).

6. A. The Normal Distribution Although all normal distributions have the same general bell-shaped curve, the exact location and shape of the curve is determined by the mean  and standard deviation  of the variable. (Notice that the normal curve is always symmetric about the vertical line x = .) examples: 1. The height of trees in a park is normally distributed with mean 10 meters and standard deviation 3 meters. 2. The time it takes Sean to get to school is normally distributed with mean 15 minutes and standard deviation 1 minute.

7. A. The Normal Distribution If a continuous variable X is normally distributed with mean  and standard deviation , we write X ~ N( ,  2 ).

8. A. The Normal Distribution

9. Example 1 The chest measurements of 18 year old male footballers are normally distributed with a mean of 95 cm and a standard deviation of 8 cm. a Find the percentage of footballers with chest measurements between: i. 87 cm and 103 cm ii. 103 cm and 111 cm b Find the probability that the chest measurement of a randomly chosen footballer is between 87 cm and 111 cm.

10. A. The Normal Distribution Solutions to example 1:

11. B. Probability using a calculator Example 2: If X ~ N(10, 2.3 2 ), find these probabilities: a. P(8 < X < 11) b. P(X < 12) c. P(X > 9). Illustrate your results.

12. B. Probability using a calculator For the Ti-84: 2 nd VARS = DISTR Choose “2” = normalcdf (“lowest bound”, “highest bound”, mean, standard deviation)

13. B. Probability using a calculator Solutions to example 2: a. normalcaf(8, 11, 10, 2.3) = 0.4759 P(8 < X < 11) = 0.4759 b. normalcaf(-1e99, 12, 10, 2.3) = 0.8077 P(8 < X < 11) = 0.4759 c. normalcaf(9, 1e99, 10, 2.3) = 0.6681 P( X > 11) = 0.6681 Note: for continuous distribution P(X > 9) = P(X < 9)

14. B. Probability using a calculator Example 3: In 1972 the heights of rugby players were approximately normally distributed with mean 179 cm and standard deviation 7 cm. Find the probability that a randomly selected player in 1972 was: a. at least 175 cm tall b. between 170 cm and 190 cm.

15. B. Probability using a calculator Solutions to example 3: If X is the height of a player then X is normal distributed with  = 179,  = 7. a.b.

16. C. Quantiles or k-Values To understand the definition of a quantile, we need to look at an example: Consider a population of crabs where the length of a shell, X mm, is normally distributed with mean 70 mm and standard deviation 10 mm. A biologist wants to protect the population by allowing only the largest 5% of crabs to be harvested. He therefore asks the question: “95% of the crabs have lengths less than what?”. To answer this question we need to find k such that P(X < k) = 0.95. The number k is known as a quantile, and in this case the 95% quantile.

17. C. Quantiles or k-Values When finding quantiles we are given a probability and are asked to calculate the corresponding measurement. This is the inverse of finding probabilities, and we use the inverse normal function on our calculator. invNorm(quantile, mean, standard deviation)

18. C. Quantiles or k-Values Example 4: If X ~ N(23.6, 3.1 2 ), find k for which P(X < k) = 0.95

19. C. Quantiles or k-Values Solution to example 4: m = 23.6 and s = 3.1 invNorm(0.95, 23.6, 3.1) = 28.69904624 Therefore P(X < 28.7) = 0.95

20. C. Quantiles or k-Values To deal with P(X > k) = p, we use P(X < k) = 1 – p

21. C. Quantiles or k-Values Example 5: A university professor determines that 80% of this year’s History candidates should pass the final examination. The examination results were approximately normally distributed with mean 62 and standard deviation 12. Find the lowest score necessary to pass the examination.

22. Solution to example 5: invNorm(0.2, 62, 12) = 51.9005452 k ≈ 51.9 So the minimum pass mark is 52.