Exam-style practice: Paper 3, Section A: Statistics

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1 Exam-style practice: Paper 3, Section A: Statistics a Use the cumulative binomial distribution tables, with n = 4 and p =.5. Then PX ( ) P( X ).5867 =.433 (4 s.f.). b In order for the normal approximation to be used as an approximation to the binomial distribution the two conditions are: (i) n is large (>5); and (ii) p is close to.5. c The two conditions for the normal approximation to be a valid approximation are satisfied. np and np( p) (3 s.f.). Therefore B (5,.5) N(3,7 9 ) so that P( B ) P( N.5).46 (4 s.f.). d If the engineer s claim is true, then the observed result had a less than % chance of occurring. This would mean that there would be insufficient evidence to reject her claim with a two-tailed hypothesis test at the % level. Though it does provide some doubt as to the validity of her claim. a Since A and C are mutually exclusive, PA ( C) and their intersection need not be represented on the Venn diagram. Since B and C are independent, PB ( C) P( B) P( C) Using the remaining information in the question allows for the other regions to be labelled. Therefore the completed Venn diagram should be: b PA ( ) P( B) P( A B) and so the events are not independent. P( A B' ).. c P( A B').444 (3 s.f.) PB ( ') P( C ( AB)') P( C).6 d P( C ( A B)').35 P(( A B)') P( A B).8 3 a The variable t is continuous, since it can take any value between and 6 (in degrees Celsius). b Estimated mean 9.49; estimated standard deviation.84 (3 d.p.). c Temperature is continuous and the data were given in a grouped frequency table. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

3 3 d The th percentile is 3.th value. Using linear interpolation: P 3.5 3. P 3.5.. P 3 3.5.55 3.5 4.5 6.5 3.5 8 3 6 6 The 9th percentile is 9 3 7.9th value. Using linear interpolation: P9 7.9 6 P9.9.9 P 9 3.

2 3 3 d The th percentile is 3.th value. Using linear interpolation: P P P The 9th percentile is th value. Using linear interpolation: P P9.9.9 P Therefore the th to 9th interpercentile range is e Since the meteorologist believes that there is positive correlation, the hypotheses are H : H : The sample size is 8, and so the critical value (for a one-tailed test) is.65. Since r =.6 <.65, there is not sufficient evidence to reject H, and so there is not sufficient evidence, at the 5% significance level, to say that there is a positive correlation between the daily mean air temperature and the number of hours of sunshine. 4 a The value of.9998 is very close to, indicating that the plot of x against y is very close to being a n linear relationship, and so the data should be well-modelled by an equation of the form q kt. b Rearranging the equation y.76.37x log q log t q.76.37logt.76.37logt logt q t Therefore.76 k.9 (3 s.f.) and n.37. c It would not be sensible to use the model to predict the amount of substance produced when t 85 C, since this is considerably outside the range of the provided data (extrapolation). Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

5 a P( Z a).5 a.96 and P( Z a).5 a.645. Therefore, for the given distribution, 3.46.96 and 4.858.645. Rearranging these equations: 3.46.96 3.46.96 4.858 and.645 4.858.645. Now subtract the second equation from the first to obtain: 4.

3 5 a P( Z a).5 a.96 and P( Z a).5 a.645. Therefore, for the given distribution, and Rearranging these equations: and Now subtract the second equation from the first to obtain: (3.46 ).645 (.96 ) and so, using the first equation, Using these values within the normal distribution, P(3.5 X 4.6) P(4.6) P(3.5) (4 s.f.) of the cats will be of the standard weight. b Using the binomial distribution, PB ( ) PB ( 9).594 =.946 (4 s.f.). c Assume the mean is 4.5kg and standard deviation is.5. Then the sample X should be normally distributed with X ~ N 4.5,.5. The hypothesis test should determine whether it is statistically significant, at the % level, that the mean is not 4.5kg. Therefore the test should be - tailed with H : 4.5 H : 4.5 The critical region therefore consists of values greater than a where P( X a).5 and so a 4.74 (4 s.f.) and values less than b where P( X b).5 and so b 4.58 (4 s.f.). Since the observed mean is 4.73 and 4.73 < a = 4.74, there is not enough evidence, at the % significance level, to reject H i.e. there is not sufficient evidence to say, at the % level, that the mean weight of all cats in the town is different from 4.5kg. 6 It is first worth displaying the information in a tree diagram. Let J denote the event that Jemima wins a game of tennis and J ' be the event that Jemima loses a game of tennis. Since Jemima either wins or loses each game of tennis, P( J) P( J '). This allows for the other probabilities on the tree diagram to be filled in. Therefore the completed tree diagram should be: The required probability is then: P( wins both games wins second game) P(wins both games) (3 s.f.) P(wins second game) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 Exam-style practice: Paper 3, Section B: Mechanics 7 r vdt ( 6 t ) i tjdt 6 3 t t t i j c 3 At t = s, r = 5i m 5i i j c c 5i j r t t 3 5i t j When t = 3 s, r i 9 j r 43i 4j s r At t = 3 s, P is 43. m from O (3 s.f.). 8 R : R : u x o cos u co 3 y o s 5 a R : u y = 5 ms, s = m, a = g = 9.8 ms, t =? s ut at 5t4.9t 4.9t 5t The solution t = corresponds to the time the arrow is fired and can therefore be ignored. 5 t The arrow reaches the ground after. s (3 s.f.). b At maximum height, v y = R : u y = 5 ms, v y = m, a = g = 9.8 ms, s =? v u as s 5 s 5 s The maximum height reached by the arrow is 8 m (3s.f.). Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

5 8 c At t = 3 s, R : v x = u 5 3 ms since horizontal speed remains constant. x R : u y = 5 ms, t = 3 s, a = g = 9.8 ms, v y =? v u at v y 5 (39.8).6 The speed at t = 3 s is given by: v v v v x y v The speed of the arrow after 3 s is 89. ms (3 s.f). 9 a u = i ms, t = s, a.i.8j ms, r =? r ut at r i (.i.8 j) r i i 4j After s, the position vector of the cyclist is 3i 4j m. b s r s After s, the cyclist is 5 m from A. c For t > s, v = 5i ms and a = The position vector is now given by: r 3i 4j + v ( t ) i r 3i 4j 5( t ) i r 5t i 4j The cyclist will be south-east of A when the coefficient of i is positive and coefficient of j is negative, but both have equal magnitude. 5t 4 5t 6 6 t 5 The cyclist is directly south-east of A after s. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

9 d First, work out the position vector of B from A: r 5t i 4j Cyclist reaches B when t = + 3 = 4 s r 5 4 i 4j r 9i 4j Let θ be the acute angle between the horizontal and B (as shown in the diagram).

6 The tension in the string immediately after the particles begin to move is 8.6 N. b Considering P: Resolving vertically R 3g Resolving horizontally and using Newton s second law of motion with a =.

6 9 d First, work out the position vector of B from A: r 5t i 4j Cyclist reaches B when t = + 3 = 4 s r 5 4 i 4j r 9i 4j Let θ be the acute angle between the horizontal and B (as shown in the diagram). 4 Then tan To the nearest degree, the bearing of B from A is 9 + = o. a Considering Q and using Newton s second law of motion: a =.5 ms, m = kg F ma gt.5 T The tension in the string immediately after the particles begin to move is 8.6 N. b Considering P: Resolving vertically R 3g Resolving horizontally and using Newton s second law of motion with a =.5 ms and m = 3 kg: T R 3.5 3g T The coefficient of friction is.58 (3 s.f.), as required. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 3

7 c Consider P before string breaks: u = ms, t = s, a =.5 ms, v =? v u at v.5 After string breaks, the only force acting on P is a frictional force of magnitude F R 3g Using Newton s Second Law for P, F ma 3g 3a a g a The acceleration is in the opposite direction to the motion of P, hence u = ms, v = ms, a =.5 ms, t =? v u at 5.7t t P takes.75 s (3 s.f.) to come to rest. d The information that the string is inextensible has been used in assuming that the tension is the same in all parts of the string and that the acceleration of P and Q are identical while they are connected. a The rod is in equilibrium so resultant force and moment are both zero. 5 5 tan sin 3 and cos 3 Taking moments about B: l mg T sin l mg T sin mg 3mg T as required. 5 3 b Resolving horizontally: R Tcos 3mg 6mg R 3 5 Resolving vertically: T sin R mg 3mg 5 6mg mg The coefficient of friction between the rod and the wall is 5. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 4

b (3 s.f.). b Make sure your hypotheses are clearly written using the parameter ρ:

b (3 s.f.). b Make sure your hypotheses are clearly written using the parameter ρ: Review exercise 1 1 a Produce a table for the values of log s and log t: log s.31.6532.7924.8633.959 log t.4815.67.2455.3324.4698 which produces r =.9992 b Since r is very close to 1, this indicates that