Statistics 1 Revision Notes - Mr Barton Maths

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(a) Give two reasons why this is a sensible suggestion. (b) Explain briefly why mathematical models can help to improve our understanding of realworld problems. Answer (a) Male heights cluster round a central height of approx 177/178 cm Height is a continuous random variable. Most male heights lie within 177 ± 3× 6.4 (b) Simplifies real world problems Enable us to gain some understanding of real world problems more quickly/cheaply. Question 2 Explain why the normal distribution may not be suitable to model the number of minutes that motorists are delayed by these roadworks. Answer For this data skewness is 3.9, whereas a normal distribution is symmetrical and has no skew. Question 3 Describe two features of the Normal distribution Answer Bell shaped curve; symmetrical about the mean; 95% of data lies within 2 s.d. of mean; etc. (any 2). Question 4 Give a reason to support the use of a normal distribution in this case. Answer Since mean and median are similar (or equal or very close), the distribution is (nearly) symmetrical and a normal distribution may be suitable. Allow mean or median close to mode/modal class ⇒ ), the distribution is (nearly) symmetrical and a normal distribution may be suitable. 48 14/04/2013 <strong>Statistics</strong> 1 SDB
12 Appendix –1 ≤ P.M.C.C. ≤ 1 Cauchy-Schwartz inequality Consider (a 2 1 + a 2 2 )(b 2 1 + b 2 2 ) – (a 1 b 1 + a 2 b 2 ) 2 = a 2 1 b 2 1 + a 2 1 b 2 2 + a 2 2 b 2 1 + a 2 2 b 2 2 – a 2 1 b 2 1 – 2a 1 b 1 a 2 b 2 – a 2 2 2 b 2 = a 2 1 b 2 2 – 2a 1 b 1 a 2 b 2 + a 2 2 2 b 1 = (a 1 b 2 – a 2 b 1 ) 2 ≥ 0 ⇒ (a 2 1 + a 2 2 )(b 2 1 + b 2 2 ) – (a 1 b 1 + a 2 b 2 ) 2 ≥ 0 ⇒ (a 1 b 1 + a 2 b 2 ) 2 ≤ (a 2 1 + a 2 2 )(b 2 1 + b 2 2 ) This proof can be generalised to show that (a 1 b 1 + a 2 b 2 + … + a n b n ) 2 ≤ (a 2 1 + a 2 2 + … + a 2 n )(b 2 1 + b 2 2 + … + b 2 n ) or P.M.C.C. between –1 and +1 In the above proof, take a i = (x i – ), and b i = (y i – ) ⇒ S xy = Σ (x i – )(y i – ) = Σ a i b i S xx = Σ (x i – ) 2 2 = Σ a i and S yy = Σ (y i – ) 2 2 = Σ b i P.M.C.C. = r = ⇒ r 2 ∑ = = ∑ ∑ ≤ 1 using the Cauchy-Schwartz inequality 14/04/2013 <strong>Statistics</strong> 1 SDB 49
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 and 40: The general normal distribution N(
Page 41 and 42: Example: The weights of chocolate b
Page 43 and 44: Question 2 Statistical models can b
Page 45 and 46: Question 2 Give two other reasons w
Page 47: Question 5 Interpret the value of a
Page 51 and 52: Normal Distribution, Z = The stan

Statistics 1 Revision Notes - Mr Barton Maths

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