Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

Revision For Tracking Test 5 1) What are the conditions for a Binomial distribution? 2) What are the conditions for a Poisson distribution? 3) In what circumstances can you approximate a binomial distribution with a Poisson distribution? 4) An archer fires arrows at a target and for each arrow, independently of all the others, the 1 probability that it hits the bull’s eye is 8 a. Given that the archer fires 5 arrows, find the probability that fewer than 2 arrows hit the bulls’ eye. b. The archer fires 5 arrows, collects them from a target and fires all 5 again. Find the probability that on both occasions fewer than 2 hit the bull’s eye. 5) A garden centre states that 95% of its daffodil bulbs will produce flowers the following season. Gill buys 100 bulbs. Use a suitable approximation to estimate the probability that at least 96 of the bulbs will flower next season. 6) A spinner is designed to land on red 10% of the time. Use suitable approximations to estimate the probability of a. Fewer than 4 reds in 60 turns of the spinner b. More than 20 reds in 150 turns of the spinner 7) The random variable X has a probability density function given by f(x) = 𝑘𝑥 2 0≤𝑥≤2 f(x) = 0 otherwise where k is a positive constant 3 a) Show that 𝑘 = 8 b) Calculate E(X) c) Specify fully the cumulative distribution function of X d) Find the value of the median e) Find the value of the mode 8) The diagram shows the graph of y=f(x) Sketch on separate diagrams, the graphs of a. y = |f(x)| b. y = f(|x|) c. y = f(x-2) d. y = -f(x) e. y = f(x) – 2 f. y = f(-x) g. y = f(2x) On each sketch (apart from (e)), show the co-ordinates of the points where the graph crosses the x-axis 9) (a) Express 3 cos θ + 4 sin θ in the form R cos (θ – α), where R and α are constants, R > 0 and 0 < α < 90°. (b) Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ for which this maximum occurs. The temperature, f(t), of a warehouse is modelled using the equation f (t) = 10 + 3 cos (15t)° + 4 sin (15t)°, where t is the time in hours from midday and 0 t < 24. (c) Calculate the minimum temperature of the warehouse as given by this model. (d) Find the value of t when this minimum temperature occurs. 5 1 10) I = dx . 2 4 ( x 1) (a) Given that y = 1 , copy and complete the table below with values of y 4 ( x 1) corresponding to x = 3 and x = 5 . Give your values to 4 decimal places. x 2 y 0.2 3 4 5 0.1745 (b) Use the trapezium rule, with all of the values of y in the completed table, to obtain an estimate of I, giving your answer to 3 decimal places. (c) Using the substitution x = (u − 4)2 + 1, or otherwise, and integrating, find the exact value of I. 11) The points A and B have position vectors 2i + 6j – k and 3i + 4j + k respectively. The line l1 passes through the points A and B. (a) (b) Find the vector AB . Find a vector equation for the line l1 . A second line l 2 passes through the origin and is parallel to the vector i + k. The line l1 meets the line l 2 at the point C. (c) (d) Find the acute angle between l1 and l 2 . Find the position vector of the point C. 𝑑𝑦 𝜋 12) Find the particular solution of the differential equation 𝑑𝑥 = 2 𝑐𝑜𝑠 2 𝑦 𝑐𝑜𝑠 2 𝑥; 𝑦 = 4 , 𝑥 = 0 Answers 1) i) There are a fixed number of trials ii) Each trial results in success or failure iii) The trails are independent iv) The probability of success at each trial is constant 2) i) Events occur singly in space or time ii) events are independent of each other iii) events occur at a constant rate 3) When n is large and p is small 4) a) 0.879 b) 0.773 5) Y = the number of bulbs that do not flower. Y~B(100,0.05) YPo(5) Answer = 0.4405 6) a) X~Po(6) Answer = 0.1512 b) X~N(15,13.5) Answer = 0.0668 7) a) 𝑘 = 3 8 b) 1.5 c) F(x) = 0 (x<0) F(x) = 𝑥3 8 (0 ≤ 𝑥 ≤ 2) F(x) =1 (x>2) d) 1.59 e) 2 9) (a) 5 cos(-53.1) (b) max value = 5 where 53.13 (c) 5○ (d) t = 15.5 10) (a) 0.1847, 0.1667 (b) 0.543 11) a) i – 2j + 2k (c) 2 + 8 ln 5 6 b) 2i +6j – k + ( i – 2j + 2k) c) 45 d) 5i + 5k 12) tan y = ½ sin 2x + x + 1