Statistics 1 Revision Notes - Mr Barton Maths

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We need the value of Z such that Φ(Z) = 0⋅01 From the tables Z = –2⋅3263 to 4 D.P. from tables (remember to look in the small table after the Normal tables) Standardising the variable ⇒ Z = X − μ σ = t − 90 15 t − 90 ⇒ = − 2 ⋅3263 to 4 D.P. from tables 15 ⇒ t = 55⋅1 to 3 S.F. 0⋅01 z 0 so the manufacturer should give a guarantee period of 55 months (4 years 7 months) Example: The results of an examination were Normally distributed. 10% of the candidates had more than 70 marks and 20% had fewer than 35 marks. Find the mean and standard deviation of the marks. Solution: First we need the values from the tables ⇒ Φ(–0⋅8416) = 0⋅2, and 1 – Φ(1⋅2816) = 0⋅1 0⋅2 0⋅1 x −0⋅8416 0 1⋅2816 Using Z − μ = X we have σ – 0⋅8416 = 35 − μ σ ⇒ μ = 35 + 0⋅8416σ − μ and 1⋅2816 = 70 σ ⇒ μ = 70 – 1⋅2816σ 0⋅2 35 μ 70 0⋅1 x ⇒ σ = 16⋅5 and μ = 48⋅9 to 3 S.F. simultaneous equations 40 14/04/2013 <strong>Statistics</strong> 1 SDB
Example: The weights of chocolate bars are normally distributed with mean 205 g and standard deviation 2⋅6 g. The stated weight of each bar is 200 g. (a) Find the probability that a single bar is underweight. (b) Four bars are chosen at random. Find the probability that fewer than two bars are underweight. Solution: (a) Let W be the weight of a chocolate bar, W ~ N(205, 2⋅6 2 ). Z = = · = – 1⋅9230769… P(W < 200) = P(Z < – 1⋅92) = 1 – Φ(1⋅92) = 1 – 0⋅9726 ⇒ probability of an underweight bar is 0⋅0274. (b) We want the probability that 0 or 1 bars chosen from 4 are underweight. Let U be underweight and C be correct weight. P(1 underweight) = P(CCCU) + P(CCUC) + P(CUCC) + P(UCCC) = 4 × 0⋅0274 × 0⋅9726 3 = 0⋅1008354753 P(0 underweight) = 0⋅9276 4 = 0⋅7403600224 ⇒ the probability that fewer than two bars are underweight = 0⋅841 to 3 S.F. 14/04/2013 <strong>Statistics</strong> 1 SDB 41
Page 1 and 2: Statistics 1 Revision Notes June 20
Page 3 and 4: 4 Median (Q 2 ), quartiles (Q 1 , Q
Page 5 and 6: 11 Context questions and answers 42
Page 7 and 8: 2 Representation of sample data Var
Page 9 and 10: Cumulative frequency curves for gro
Page 11 and 12: 3 Mode, mean (and median) Mode The
Page 13 and 14: 5 20 since 1 and Exa
Page 15 and 16: The interquartile range, I.Q.R., is
Page 17 and 18: ⇒ Q 1 = 4⋅5 + 5 × = 7⋅7421
Page 19 and 20: The same ideas apply for a continuo
Page 21 and 22: 1 2 1 Rough chec
Page 23 and 24: 6 Probability Relative frequency Af
Page 25 and 26: Tree diagrams The rules for tree di
Page 27 and 28: But ⇒ ⇒ P(A ∩ B) = P(A) × P(
Page 29 and 30: Coding and the PMCC To see the effe
Page 31 and 32: (c) r = - 0⋅725 is ‘quite close
Page 33 and 34: ⇒ b = = . . = 1⋅773584906
Page 35 and 36: Expectation algebra E[aX + b] =
Page 37 and 38: The expected variance, Var[X] = σ
Page 39: The general normal distribution N(
Page 43 and 44: Question 2 Statistical models can b
Page 45 and 46: Question 2 Give two other reasons w
Page 47 and 48: Question 5 Interpret the value of a
Page 49 and 50: 12 Appendix -1 ≤ P.M.C.C. ≤ 1 C
Page 51 and 52: Normal Distribution, Z = The stan

Statistics 1 Revision Notes - Mr Barton Maths

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