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. Standard Normal Distribution . What if we don’t have a ‘nice’ number of standard deviations from the mean? Eg: A population of gummy bears has a mean of 3.4kg and standard deviation of 0.5kg What % are less than 4.1kg? How can we compare distributions with different means and standard deviations? Use STANDARD normal distribution Standard deviation = 1 Mean = 0 z xx 34 kg 0 z Standard deviation 0 7 kg x mean 25 kg x z = the number of standard deviations from the mean . . Z values Theta Ex 24.04 . z = the number of standard deviations from the mean For all z values (to 2dp) a table lists the areas under the standard normal curve This gives the corresponding probability for each z value. The probability is for the region from z back to the middle (0) Look up the z value to get the probability 0 z P(0 < z < 1.2) = 0.3849 P(0 < z < 1.5) = P(0 < z < 0.8) = P(0 < z < 2.1) = P(z < 1.2) = P( z < 1.8) = P(z > 0.6) = 0 1.2 P(z < -1.2) = -1.2 0 0 0.6 P( z < -2.3) = P(-1.4 < z < 1.8) = -1.4 0 1.8 . Converting To Standard Normal . A population of snails have a mean weight of 54g (standard deviation of 8g) (a) Find the probability a snail is over 60g Steps: 1) Draw a distribution curve 54g 60g 2) Convert to a standard normal distribution 3) Look up ‘z’ to get probability 4) Add / subtract 0.5? 0 (b) Find the probability a snail is under 68g WORKING 1) Question 2) ‘Z’ value 3) Probability 4) Answer 54g z= . Normal Distribution Example . Theta Ex24.06 A fish farm has salmon with mean weight of 880g and standard deviation of 160g a) What is the probability a fish is under 800g? b) What % of fish are under 1kg? c) From a population of 2600 fish, how many fish would be between 600g and 1200g? . Normal Distribution: Practice . Annual sunshine hours in Nelson are normally distributed with a mean of 2200hr and a standard deviation of 35hr. 1) In a randomly chosen year: a) What is the probability that the annual sunshine hours in Nelson are over 2150hr? b) What is the chance that the annual sunshine hours in Nelson is less than 2100hr? 2) For how many of the past 100 years would we expect the number of sunshine hours to have been between 2150 and 2250hr? 1a) The z-value for 2150 is given by z 2150 2200 1.429 35 Working 1) 2) 3) 4) Question ‘Z’ value Probability Answer P(x > 2150) = P( z > - 1.429) = 0.5 + 0.4235 = 0.9235 1b) The z-vale for 2100 is given by z 2150 -1.429 2200 0 2235 The probability the annual sunshine hours are over 2150 h is 0.9235. 2100 2200 2.857 35 1) 2) 3) 4) Question ‘Z’ value Probability Answer P(x < 2100) = P( z < - 2.857) 2100 2200 = 0.5 – 0.4984 -2.857 0 = 0.0016 The probability the annual sunshine hours are less than 2100 h is 0.0016. 2. 1) 2) 3) 4) The z-values for 2150 h and 2250 h are -1.429 and 1.429 Question P(2150< x < 2250) ‘Z’ value = P( -1.429 < x < 1.429) Probability = 0.4235 + 0.4235 Answer = 0.8470 2150 -1.429 2200 0 2250 1.429 Therefore over the past 100 years we would expect there to have been 84.7 or 85 years with between 2150 and 2250 h of sunshine.