1. Normal distribution Unit 8 strand 1

4. symmetrical about the mean

5. it extends from –infinity to + infinity

7. Approximately 99.9% of the distribution lies between 3 SDs of the mean

8. The probability that X lies between a and b is written as: P(a<X<b). To find the probability we need to find the area under the normal curvebetween a and b. We can integrate or use tables.

9. The probability that X lies between a and b is written as: P(a<X<b). To find the probability we need to find the area under the normal curvebetween a and b. We can integrate or use tables. To simplify the tables for all possible values of mean and s² the variable X is standardised so that the mean is zero and the standard deviation is 1. The standardised normal variable is Z and Z (0, 1)

11. So, the probability of Z < 0.16 is 0.5636.

12. Find the probability of Z < 0.429

13. Now try these Draw sketches to illustrate your answers If Z ~N (0, 1), find 1. P (Z <0.87) 2. P (Z > 0.87) 3. P (Z < 0.544) 4. P (Z > 0.544)

16. Example Find P (Z < 0.411) P (Z > - 0.411) P (Z > 0.411) P (Z < - 0.411) Solution P(Z < 0.411) = [from tables 6591 + 4] = 0.6595 P(Z > - 0.411) = P(Z < 0.411) = 0.6595 (from (above)) P(Z > 0.411) = 1 – Φ(0.411) = 1 – 0.6595 = 0.3405 P(Z < - 0.411) = P(Z > 0.411) = 0.3405

17. Now try these 1. P (Z > - 0.314) 2. P (Z < - 0.314) 3. P (Z > 0.111) 4. P (Z > - 0.111)

18. What if we don’t have a distribution where the mean is 0 and the standard deviation is 1?

19. Standardising a normal distribution To standardise X where X ~ N(μ, σ²) subtract the mean and then divide by the standard deviation: where Z ~ N(0,1)

24. Task 3 A forensic anthropologist has asked your advice. She was investigating the lifespan of insects on a human corpse. The mean lifespan for one insect is 144 days and the standard deviation is 16 days. Find the probability that one insect will live less than 140 days and another more than 156 days. (M1, D1)