2. If you observed the heights of people at a busy airport you will notice that some are short, some are tall, but most would be around a central value (which would probably depend on their Gender). If you displayed a distribution of heights for men and women, you would get something like: 172cm 178cm Female Heights Male Heights

3. Height is an example of a CONTINUOUS RANDOM VARIABLE. Height is a particular type of random variable whose distribution Is symmetrical in shape – a sort of bell shape. This sort of distribution is called the NORMAL DISTRIBUTION The characteristics of the normal distribution are:  Symmetrical about the mean, μ  Mode, median and mean are equal due to the symmetry of the distribution  The range of X is from -∞ to +∞  The horizontal axis is asymptotic  The total area under the curve is 1  2 / 3 of the values lie within 1σ of μ  95% lie within 2σ of μ  99.7% lie within 3σ of μ μ

4. How does changing μ and σ affect the normal distribution? μ1μ1 μ2μ2 μ3μ3 μ1μ1 μ 2 < < μ 3 Changing the mean translates the curve along the x-axis. It DOES NOT change its shape.

5. How does changing μ and σ affect the normal distribution? μ σ1σ1 σ 2 < < σ 3 Increasing σ makes the curve fatter and shorter Decreasing σ makes the curve thinner and taller. It DOES NOT MOVE.

6. If a particular random variable, X, has a normal distribution, We define the distribution using the notation: X ~ N(μ, σ 2 ) If we are finding P(a < X < b), you are finding an area underneath the curve a b mean variance

7. STANDARD NORMAL DISTRIBUTION Situations that demonstrate a Normal distribution will demonstrate different means and variances and so we need a standardised model that we can transform each situation to, to allow us to perform calculations with ease. For this, we use the STANDARD NORMAL DISTRIBUTION. It has the properties of any other normal distribution but Has a mean of 0 and a standard deviation (and variance) of 1. We give this particular normal distribution the letter Z. So: Z ~ N(0, 1 2 ) Probabilities for this distribution are provided in statistical tables.

8. Any normal distribution can be transformed onto the standard normal distribution using the following transformation: So, if we have the model X ~ N(10, 4 2 ) and want to find P(12<X<16) We can transform the values from X to values on Z by: In other words, P(12 < X < 16) on X~ N(10, 4 2 ) would be P(0.5 < Z < 1.5) on Z~N(0, 1 2 ) = P(0.5 < Z < 1.5)

9. 12 16 0.5 1.5 0.5 X ~ N(10, 4 2 ) Z ~ N(0, 1 2 ) P(12 < X < 16) P(0.5 < Z < 1.5) P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5) = 0.9332- 0.6915 = 0.2417

10. If X ~ N(10, 4 2 ), what is P(X < 8)? = P(Z < - 0.5) - 0.5 0.5 0.6915 P(X < 8) = 1 – 0.6915 = 0.3085 0.5

11. Ex 9A Qu 6 The random variable, A is defined by A ~ N(28, 3 2 ). Find: (a)P(A 36)(c) P(28 < A < 35) (d) P(22 < A < 26)(e) P(25 < A < 33) (a)P(A < 32) = = P(Z < 1.33) = 0.9082 (b) P(A > 36) = = P(Z > 2.67) = 1 – 0.9962 = 0.0038 2.67

12. (c) P(28 < A < 35) = = P(0 < Z < 2.33) = 0.9901 – 0.5 = 0.4901 (d) P(22 < A < 26) = = P(-2 < Z < -0.67) = (1-0.7486) – (1-0.9772) = 0.2286 (e) P(25 < A < 33) = = P(-1 < Z < 1.67) = 0.9525 – (1-0.8413) = 0.7938

13. Ex 9A Qu 13 A machine dispenses liquid into cartons in such a way that the amount of liquid dispensed on each occasion is normally distributed with standard deviation 20ml and mean 266ml. Cartons that weigh less than 260ml have to be recycled. What proportion of cartons are recycled? Cartons weighing more than 300ml produce no profit. What percentage of cartons will this be? X ~ N(266, 20 2 )

14. Cartons that weigh less than 260ml have to be recycled. What proportion have to be recycled? P(X < 260) = = P(Z < -0.3) = 1 – 0.6179= 0.3821 = 38.21% Cartons weighing more than 300ml produce no profit. What percentage of cartons will this be? P(X > 300) = = P(Z > 1.7) = 1 – 0.9554= 0.0446 = 4.46%

15. Ex 9A Qu 15 Over a long period it has been found that the breaking strains of cables produced by a factory are normally distributed with mean 6000N and standard deviation 150N. Find, to 3 decimal places, the probability that a cable chosen at random from the production will have a breaking strain of more than 6200N. X ~ N(6000, 150 2 ) P(X > 6200) = = P(Z > 1.33) = 1 – 0.9082= 0.0918 = 0.092 to 3dp

16. Ex 9A Qu 18 The thickness of some sheets of wood follows a normal distribution with mean μ and standard deviation σ. 96% of the sheets will go through an 8mm gauge while only 1.7% will go through a 7mm gauge. Find μ and σ. 8 μ -(1)

17. 7 μ -(2) -(1) -(2) (1)-(2) 1 = 3.87σ σ = 0.258mm Sub σ = 0.258 into (1) 8 – μ = 1.75 x 0.258 μ = 7.5485mm

18. Packets of breakfast cereal are said to contain 550g. The manufacturer knows that the weights are normally distributed with mean 551.2g and standard deviation 15g. What proportion of packets will contain more than the stated weight? X ~ N(551.2, 15 2 ) P(X > 550) = = P(Z > -0.08) = 0.5319= 53.19%

19. A biologist has studied a particular tropical insect and she has discovered that its life span is normally distributed. The mean lifespan of this insect is 72 days and the standard deviation of its lifespan is eight days. Find the probability that the next insect studied lives: (a)Fewer than 70 days(b) More than 76 days (c) Between 68 and 78 days. X ~ N(72, 8 2 ) (a)P(X < 70) == P(Z < -0.25) = 1 - 0.5987= 0.4013

20. X ~ N(72, 8 2 ) (b)P(X > 76) == P(Z > 0.5) = 1 - 0.6915= 0.3085 (c)P(68 < X < 78) = = P(-0.5 < Z < 0.75) = 0.7734 – (1 – 0.6915) = 0.4649

21. The lengths of metal bolts are normally distributed with mean μ and standard deviation 7cm. If 2.5% of the bolts measure more than 68cm, find the value of μ. X ~ N(μ, 7 2 ) P(X > 68) = 0.025 P(Z > 1.96) = 0.025

22. Jam is packed in tins of nominal net weight 1kg. The actual weight of jam delivered to a tin by the filling machine is normally distributed about the mean weight set on the machine with a standard deviation of 12g. The average filling of jam is 1kg. (a) Calculate the probability that a tin chosen at random contains less than 985g. X ~ N(1000, 12 2 ) (a)P(X < 985) == P(Z < -1.25) = 1 - 0.8944= 0.1056

23. It is a legal requirement that no more than 1% of tins contain less than the nominal weight. (b)Calculate the minimum setting of the filling machine which will meet this requirement. X ~ N(μ, 12 2 ) (b)P(X < 1000) = 0.01 P(Z > 2.3263) = 0.01P(Z < -2.3263) = 0.01