Normal Distribution - A Level

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you can oc: The escact shape of will 6²52 normal distribution can be 1 0√√20 This distribution can be described as NCU, 0²) and the standard deviation two parameters. the mean the axis measurements measure from a a ý xi becomes yi, you -122² e Standardised variable Z has mean O, variance 1 is N (0, 1) distribution After the ranable. JC has been transformed to 2 2 = (00-μ)/0 the form of the now usundardised Curve is -12 12 (x-μ)² u the the curve isn't as useful as Modelling discrete situations. The normal distribution applies strictly to a continuous variable. However, it can also be used for a discrete variable providing that.... • The distribution is approximately normou (meaning that the steps in the in comparison to the are small deviation). • The continuity corrections are applied where needed. Continuity corrections need to be carefully considered. you add/Subtract correctly. So If the region is given as an inequality then.... 20≤x≤ 30 becomes 19.5 30.5 20<x< 30 becomes 20.5<< 29.5 Approximating the binomial distribution! can the normal distribution as an binomial approximation for the •n is large •np is not too close to 0 or n These are combined 0« np <<n The parameters for the normal distribution are mean: ме пр vanance: ở npCI-D) ng which can also be clenoted as You Positive skew use = Possible valves. standard as the conditions negative shew distribution if. 1610 N(np, npq) symmetrical Normal Distribution XC~ N (mean, vanance) X-N (μ₁ 0²) if if lif -3 s.d. X 7 x = 68% g :95% of values 99.7% of values & A 2 the lack of value. if The two MX4 x < You can I ( T 1 1 ( u values T 1 198 then then values lie within 1 S.d. of the mean lie within 2 S.d. the lie within 3 S.d. 2 then then we then use X<18 the or you. can everything that mean, median and mode are all equal s.d. use Your calculater only takes o So remember to noot the value. 3 CD normal PD normal CD normal then value = 18. use normal CD values you've been given. and either use lower = doesn't matter you just use the exact valve = 18 and upper 29/1/21 X7 then Solve them use of ģÝ the ११ upper = 9 x 10⁹⁰ mean mean separately Lower = -9x1099 and input normal CD.! and add and upper (which finds isn't in the regions) so you can valve. Interpreting sample data using If a on population has a standard deviation of o can dedore it by normally is distributed, we X~ N(₂0²) If we then took a sample from that population, Sample is. Still normally distributed and the mean will still deviation will now be n=no. in sample The difference between the P(x>4.04) 19 u Smaller & = 1- p(x< 4.04) = 1-0.691462.. smaller. = 0.3085375383 P=0.3085 the normal when 0²/n = new, smaller Standard deviation. 2 can be shown by..... "NCM, 0²/n) 2612121 N(₂₂0²) Example 1: A machine is designed to make paperclips with a mean deviation of 0.089. The distribution of the masses is normal. MAD il find the probability that an individual paperclip chosen mass greater than 4.049 ~NC4, 0.08²) ● Lower - - 9 x 10 upper = 4.04 = 0·08 μ = 4 ११ ii find the sample probability = 1 M that the mean 25 paperclips N(4, 0.08²) P(x > 4.04) 25 P(x ≤4.04) → lower = - 9 × 1099 upper = 4.04 u = 4 J = 0·0815 0.99379... 0.0062096654 M=4 Area 0.95 0 = 0·08/5 u = 4 = 0.00621 ли 2$ A quality control officer weighs a random sample of Paperclips and finds their mass to be 101.2.91 a hypothesis test at the 5% sig level to find out whether this produces evidence that there has been increase in the mean mass. an H₂: M= 49 H₁ μ> 49 To find the Critical value.….... is greater than 51 mass of a random 84.049 } ← 0.05 mean of the Officers sample= ·04 89 101. square root the 0.08²/25 • This is the region where to is rejected You need to use the inverse normal! = 4.02 632 < Critical value. Since 4.0487 4.026 it is within the Critical (shaded) Conclude that to can be region. Therefore, we can rejected as there is sufficient evidence that there has been an increase in ³ mass. The hypothesis test can also be carried out using 2- valuies. Hypothesis testing using 2 values: for question ini, NCO, 1²) So the 2 value for the top 5%. is.... area=0.95 1.644 85 3667 using inverse normal μ = 0 mean mass of the Officers sample: 101.2/25 = value of this.... So to find the Z = 4.048- 4 (0.08/5) yout 2 value = Since 371.645, we value lies with in can reject Ho because the z Critical region. we can conclude that the test is significant. known and estimated standard deviation To complete a hypothesis tut you nethe values (>50) youen of the entire 18 Х Observed value - expected value Standard deviation = 3 S The variance of the sample is usually s² so that the variance of entire population ist You will need to remember that. MX n need us and o but if entire population, have a sufficiently large sample population you can estimate the standard deviation population. =4.0489 and S= ≤x²-no² Example 2: The mass of a breed mean of 72.79 and The weights of 12 unidentified lizards are: 80.4 75.51 i State the hypotheses and if it's a H₂: 67.2174.9178.8176.51 80.2181.9179.3170169-2169.11 1 or 2 tailed test. = 72.7 u= H₁: 72.7 →it's a 2 tailed test because the lizards Could weigh more or less than the known breed. ii Carry out a hypothesis test at the 5% significance level. decide if the unknown lizards are the same breed as the known ones. To carry out a hypothesis test, we need the mean of the sample.. u of sample lizard is normally distributed with a bottom 2-5%. Set up your model: 4.82 X-N (72.7, find the 2 Critical regions... Area 0.025 4.8/√T2 72.7 O u = = 69.98 total weights no. of lizards A K (use inverse normall u Area = 1-0·025 0 = 4.815T2 M=72-7 = 75.42 = 903 12 75.25 mean of region so are different breeds of lizard. Top 2.5% 75.25 75.42 therefore, we accept Ho because the the sample population lies within the acceptance there is insufficient evidence to suggest that they