1. 381 Continuous Probability Distributions (The Normal Distribution-I) QSCI 381 – Lecture 16 (Larson and Farber, Sect 5.1-5.3)

2. 381 Introduction The (or Gaussian) distribution is probably the most used (and abused) distribution in statistics. Normal random variables are continuous (they can take any value on the real line) so the Normal distribution is an example of a continuous probability distribution.

3. 381 The Normal Distribution-I -- -2  -3  +3  +2 ++ The graph of the normal distribution is called the normal (or bell) curve. Total area=1 Inflection points

4. 381 The Normal Distribution-II The mean, median, and mode are the same. The normal curve is symmetric about its mean. The total area under the normal curve is one. The normal curve approaches, but never touches, the x-axis.

5. 381 The Normal Distribution-III Notation We say that the random variable X is normally distributed with mean  and standard deviation  The equation defining the normal curve is: f (x) is referred to as a (pdf).

6. 381 We do not talk about the probability P[X=x] for continuous random variables. Rather we talk about the probability that the random variable falls in an interval, i.e. P[x 1  X  x 2 ]. P[x 1  X  x 2 ] can be determined by finding the area under the normal curve between x 1 and x 2. Normal Distributions and Probability-I

7. 381 Normal Distributions and Probability-II 99.7% 95% 68% 34% 13.5% Of the total area under the curve about: 68% lies between  -  and  +  95% lies between  -2  and  +2  99.7% lies between  -3  and  +3  μ

8. 381 Example The mean length of a fish of age 5 is 40cm and the standard deviation is 4. What is the probability that the length of a randomly selected fish of age 5 is less than 48cm? 48cm is two standard deviations above the mean so the area to the left of 48cm is 0.5+0.475 = 0.975. This is an approximate method; we will develop a more accurate method later.

9. 381 The Standard Normal Distribution-I The normal distribution with a mean of 0 and a standard deviation of 1 is called the The standard normal distribution is closely related to the z-score:

10. 381 The Standard Normal Distribution-II Area=1 The area under the standard normal distribution from -  to x: is often listed in tables. can be computed using the EXCEL function NORMDIST.

11. 381 Example-I Find the area under the standard normal distribution between -0.12 and 1.23. We use our previous approach: P[-0.12  X  1.23] = P[X  1.23] - P[X  -0.12] In EXCEL: NORMDIST(1.23,0,1,TRUE)-NORMDIST(-0.12,0,1,TRUE) NORMDIST(x, , ,cum)

12. 381 Example-II P[X  1.23]P[X  -0.12] P[X  1.23]-P[X  -0.12]

13. 381 Further Examples How would you find the probability that: X  -0.95 X  1 0.1  X  0.5 OR 1.5  X  2 0.1  X  0.5 AND 1.5  X  2

14. 381 Generalizing-I We can transform any normal distribution into a standard normal distribution by subtracting the mean and dividing by the standard deviation.  =300;  =20;x=330  =0;  =1;z=1.5 Area=0.933 in both cases Z=(x-300)/20

15. 381 Generalizing-II To find the probability that X  Y if X is normally distributed with mean  and standard deviation . Compute the z-score: z =(Y -  )/  Calculate the area under the normal curve between -  and z using tables for the standard normal distribution. We could calculate this area directly using the EXCEL function NORMDIST.

16. 381 Examples-1 The average swimming speed of a fish population is 2m.s -1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s -1. Its swimming speed is greater than 2.5m.s -1. Its swimming speed is between 2 and 3 m.s -1.

17. 381 Examples-1 The average swimming speed of a fish population is 2m.s -1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s -1. Its swimming speed is greater than 2.5m.s -1. Its swimming speed is between 2 and 3 m.s -1. = P(z < (1-2)/.5) = P(z < -2) = 0.0228

18. 381 Examples-1 The average swimming speed of a fish population is 2m.s -1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s -1. Is swimming speed is greater than 2.5m.s -1. Its swimming speed is between 2 and 3 m.s -1. = P(z > (2.5-2)/.5) = P(z > 1) = 1 - P(z ≤ 1) = 0.159

19. 381 Examples-1 The average swimming speed of a fish population is 2m.s -1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s -1. Its swimming speed is greater than 2.5m.s -1. Its swimming speed is between 2 and 3 m.s -1. = P( (2-2)/.5 ≤ z ≤ (3-2)/0.5 ) = P(0 ≤ z ≤ 2) = P(z ≤ 2) – P(z ≤ 0) = 0.477

20. 381 Examples-1 The average swimming speed of a fish population is 2m.s -1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s -1. Its swimming speed is greater than 2.5m.s -1. Its swimming speed is between 2 and 3 m.s -1. Why is the normal distribution not entirely satisfactory for this case?

21. 381 Examples-2 In a large group of men 4% are under 160cm tall and 52% are between 160 cm and 175 cm. Assuming that the heights of men are normally distributed, what are the mean and standard deviation of the distribution?

22. 381 In a large group of men 4% are under 160cm tall and 52% are between 160 cm and 175 cm. Assuming that the heights of men are normally distributed, what are the mean and standard deviation of the distribution? P(z < (160-μ)/σ ) = 0.04 = P( z < -1.75) P( (160-μ)/σ ≤ z ≤ (175-μ)/σ ) = 0.52 0.56 = P(z ≤ (175-μ)/σ ) = P(z ≤ 0.15 ) So: -1.75 = (160-μ)/ σ 0.15 = (175- μ)/ σ μ= 174, σ= 8 Examples-2 We get these from a standard normal table

23. 381 Examples-3 An anchovy net can hold 10 t of anchovy. The mean mass of anchovy school is 8 t with a standard deviation of 2 t. What proportion of schools cannot be fully caught? Hint: we are interested in the proportion of schools greater than 10t. ie. 1- P(X ≤ 10) = 1- P(z ≤ (10-8)/2) OR: = 1 - NORMDIST(10,8,2,TRUE) = 0.159