Statistics Revision Solutions

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1 Statistics Revisio Solutios (i) H ~N (00, ) ad W ~N (7, 9 ) P ( 7.<W < 7) 0.0 (ii) H + H W ~ (0, () + (8) ) N ( H + H W > 0) P (iii) H + W ~N (7, ) P ( H + W > A) > 0.9 P( H + W < A) < 0.0 A< ivnorm(0.0, 7, ) 0. 8 ad A 0 (i) M ~N (80, ) ad W ~N(, ) max T M + M + M + W + W + W + W ~ N(80 +, + 7) P ( T > 00) 0.08 (ii) T ( M + M + M ) ( W + W + W + W ) ~ (, ' N 7) P ( T ' > 0) P ( X > b) 0. P( X < b) 0. b ivnorm(0., 0, ). P ( X < a) 0. a ivnorm(0., 0, ) 0.7 ( a, b) (, ) (i) X ~N(0, ) P ( <X < ) 0.7 (ii) Y ~B(0, 0.) P ( Y ) 0. (i) S ~N(0,. ) ad T ~N(8, ) P ( S < 0) P( T < ) 0.

2 (ii) T S ~N (8, 8.) P ( T S > ) (iii) S T ~ N(,. +.) P S T > X ~N(, ) ad Y ~N(, ) X + X Y ~ N(, ) ( X + X Y < ) P( < X + X Y < ) P P X ~N(0, ) ad ~N(, P Y ( 8) 0) T 0 PX + P i Y ~ N( , ) i j j P ( T > 80) (i) X ~N(0, 0 ) [ P( X )] P ( X < m) P( X > m) < m P ( X < m) or P ( X <m) m ivnorm(, 0, 0) 0.8 (ii) Let Y be the rv deotig the umber of orages (out of 0) each havig mass more tha m grams. The Y ~B(0, ) 0 Sice p.7< ad p < 0., Y ~ P o (.7) approximately. P( more tha orages have mass less thamgrams) P( less tha orages have mass more thamgrams) P ( Y < ) P( Y ) 0.8

3 7(i) Possible digit combiatio Probability (icludig permutatio)!!!!!!!!! Total (ii) P( umbers are idetical sum of umbers is at least ) (Note that 888 ad 999 are the oly two combiatios which satisfy the umerator, ie the umbers are idetical ad their sum is at least ) 8 (i) Number of ways to seat childre (without ay restrictios) i a circle ( )! 0! Firstly, umber of ways to arrage the boys! Secodly, umber of ways to slot the girls betwee the boys C!! Required probability!! 0! (ii) Groupig J, C ad D together as a sigle uit, umber of ways to arrage the childre such that J sat betwee C ad D ( 9 )!! 8!! (ote that C ad D ca switch sides)

4 Number of ways to arrage the childre such that J, C ad D simply sat together ( )! 8!!! 9 (ote that J, C ad D ca be freely permutated withi the uit) Required probability 8! 8! 9 (i) Su(Y) Sat(Y) Su (N) Fri (Y) Sat(N) Su(Y) Su(N) Su(Y) Sat(Y) Su (N) Fri (N) Sat(N) Su(Y) Su(N) (Y): go o that day (N): ot goig o that day P(A goes o Su) (ii) P( A will go o Fri but ot o Su) +

5 P (A will go o Fri) Hece, P (A will ot go o Su A will go o Fri) (iii) P(A goes o both Fri ad Su) Hece, P( A goes o Fri A goes o Su) (i) Ubiased estimate of populatio mea Ubiased estimate of populatio variace (ii) Sice 00 is large, by CLT, X ~N.9, approx 00 P ( X <.7) 0.9 (iii) Sice 0 is large, by CLT, T X + X X 0 ~ N( ,. 0.) approx P ( T < 70) 0.. X ~ N( µ, σ ) µ P ( X < ) 0.07 PZ < 0.07 σ µ ivnorm(0.07).7 () σ. µ P ( X >.) 0.0 P( X <.) 0.9 PZ < 0.9 σ. µ ivnorm(0.9). () σ Solvig () ad () gives σ 0.9, µ.8kg P( exactly large, exactly small, exactly betwee large ad small)

6 ! (0.0)(0.07)( ) 0.08 (i) To test: H : µ 7. H : µ 7. 0 x.9, σ.8, 0 < Usig the Z test, by the GC, p 0.09> 0. 0 H 0 is ot rejected ad there is isufficiet evidece at the % level to suggest that the compay is overstatig the mass of coffee per jar. x µ (ii) Test statistic Z < ivnorm( 0.0). 07 σ x 7. <.07 x< (iii) Sice X ~ N µ σ,, the σ X µ ~ N 0, ( X µ < 0.) P( 0.< X < 0.) P < Z < P µ The above ca be re-iterpreted as P Z < ivnorm(0.0) 70.8 (iv) Sample Variace s.8 Hece ubiased estimate of the populatio variace s (.8) 8 0 9

7 x µ Test statistic Z. σ (i) Sice ( y) ( where σ 8) x, lie o the regressio lie ad it is give that x 00, The sample mea of weights y 0.09(00) 90 (ii) Sample variace for heights ( x x) [ ] 0 ( x x) 0 () Sample variace for weights ( y y) Also, b ( x x)( y y) ( x x) 0.09 [ ] ( y y) () Substitutig () ito the above gives ( x x)( y y) 0.09( 0 ) () Substitutig () ad () ito ( x x)( y y) ( y y) d gives ( ) d 9 Equatio of lie of regressio of x o yis x c+ dy c+ 9y Sice (00. ) lies o the lie, c 00 9() ad hece x + 9y (iii) r bd 0.8 r 0. 9 A strog value for the coefficiet r deotes a high degree of similarity (coicidece) betwee both regressio lies. (iv) y 0.09(00) 90 kg (a) 0)., 0 x (x ( 0) 797( 0.) ( x 0)( y 0) 97(0.) 9. 7 r S S xx xy S yy (.) (88)

8 It appears that there is a strog egative liear associatio betwee the scores o a fitess test ad the weights of the participatig studets. (b) L a+ bw () b S S LW WW LW W L ( W) W L 0.0, W Substitutig the above quatities ito (), a L bw (8.8). Hece, equatio of regressio lie is give by L W.. X ~ N, P ( X >.) < 0.0 P( X <.) > P Z.. < > 0.99 > ivnorm(0.99)... Solvig gives > Yes, cetral limit theorem (CLT) was used i approximatig X to a ormal distributio based o the fact that the sample size (umber of observatios) was large. (a) A B 0 x x x 0 (i) P ( A)

9 (ii) Number of boys that are shortsighted 0 (this is represeted by ( A B)' ) ( 0 x ) + ( x) + ( x) + x A B ( A B ) 0 + ( ) P P (iii) ( A B ) (iv) P( A B) P ( A B) P( B) 9 (b) (i) 9 77 (ii)!! (iii) (!) shortsighted o-shortsighted shortsighted girl girl boy 7(i) Let X be the radom variable deotig the umber of days a studet is late from Moday to Wedesday. The X ~B(, 0.09) Required probability P ( X ) (0.009) (ii) ( P studet is ot late o days) ( 0.009) 0. 9 Let Y be the radom variable deotig the umber of studets (out of a class of ) who are ot late for a stretch of days. The Y ~B(, 0.9) ad P ( Y ) 0.

10 (iii) P ( at least latecomer i a class of i a give school week of days) Let Y ' be the radom variable deotig the umber of weeks (out of 80 weeks) where there is at least oe latecomer. The Y ' ~ B(80, 0.77) Sice p.> ad q.8>, Y ' ~ N(., 7.9) approx. P ( Y ' > 0) P( Y' > 0.) (cotiuity correctio has to be used) 8 (a) X ~N(0, 0. ) T X + X + X + X + X ~ (0, 0. 0.) N P ( 0.<T <.) 0. (b) Y ~N(0, 0. ) S ~ N( 0 0, 0. 0.), W ~ N( 0 0, 0. 0.) ad S W ~N(0, 0.9) P ( S < W + ) P( S W < ) 0.97 (c) 0. L ~N 0,. 0 L 0 ~ N(0,. 0 ) P ( L 0 a) 0.0 P( L 0 a) + P( L 0 a) 0. 0 Sice the distributio of ( L 0) is cetred at zero, P ( L 0 a) 0. 0 P Z a P ( Z a) 0.0 a ivnorm(0.0). a 0.09

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