Statistics 2. The Normal Distribution

Transcription

1 Chapter assessment Statistics The Normal Distribution 1. Soup tins have a capacity of 65 ml. The volume of soup, X ml, dispensed into each tin is Normally distributed with mean 610 and standard deviation. If more than 65 ml is dispensed, the tin overflows. (i) Find the probability that the volume of soup dispensed into a tin is between 600 ml and 65 ml. [4] The proportion of tins containing at least 600 ml is too low. To increase this proportion to 95%, the dispenser is adjusted in such a way as to reduce the standard deviation of X while leaving the mean unchanged. (ii) Show that the new value of the standard deviation is 6.0. [4] (iii) Show that the proportion of tins overflowing is now 0.6%. [3] Following the adjustment, 1000 randomly chosen tins are inspected. (iv) Use a suitable approximating distribution to calculate the probability that the soup overflowed on more than 10 occasions when being dispensed into the tins. [4]. The number of marks gained by candidates in a particular Statistics examination, for which the maximum mark is 60, is modelled by a Normal distribution with mean 36 and standard deviation. The marks are reported as integers. (i) Find the probability that a randomly chosen candidate scores exactly 30 marks. [4] (ii) Three candidates are chosen at random. Find the probability that just one of them gets fewer than 30 marks. [3] (iii) It is intended that the proportion of candidates receiving a grade A should be as near as possible to 0%. What is the lowest integer mark that should be awarded a grade A? [4] (iv) In a future Statistics examinations it is intended that the top 5% of candidates should gain a reported mark of at least 45. Determine the required value for the mean mark, assuming the standard deviation remains at. [4] 3. The number of arrivals per minute at a drive-in fast food outlet is modelled by a Poisson distribution with mean. During a Saturday evening, = 0.7. (i) Give reasons why the proposed Poisson distribution might be a suitable model. [1] (ii) Calculate the probability of exactly two arrivals during a one-minute interval. [] (iii) Calculate the probability of at least four arrivals during a five-minute interval. [3] MEI, 10/1/07 1/6

2 (iv) Using a suitable approximating distribution, calculate the probability that there are more than 40 arrivals between 7 pm and pm. [4] Due to capacity constraints, the management of the fast food outlet would like the probability of more than 40 arrivals in an hour to be 0.0. They ask a statistician to determine the value to which should be reduced. (v) Show that must satisfy the equation [] 60 (vi) Use the substitution u 60 to formulate a quadratic equation in u. Solve this equation to show that the value of is just less than 0.5. [3] 4. Every day, Morse attempts the crossword puzzle in his newspaper. The time taken, X minutes, to complete the crossword may be modelled by a Normal distribution with mean and standard deviation 4.5. (i) Calculate the probability that he takes (A) (B) more than 5 minutes, between 15 and 5 minutes to complete the crossword. [5] (ii) What length of time would be enough for Morse to finish the crossword on 95% of days? [3] Each day Morse takes a train to work. The journey takes 5 minutes. He starts his crossword at the beginning of his journey. (iii) Find the probability that he completes the puzzle by the end of his journey at least twice in a five-day week. [4] (iv) Morse changes his newspaper and finds that on 99% of occasions he completes the crossword during his morning train journey. Assuming that the time taken, Y minutes, to complete the crossword has the distribution N(1, ²), find the value of. [3] Total 60 marks MEI, 10/1/07 /6

3 Solutions to Chapter assessment 1. (i) X 600 Z X 65 Z 1.75 P(600 X 65) P( 1.5 Z 1.75 ) (1.75 ) ( 1.5 ) (1.75 ) 1 (1.5 ) (ii) We want P( X 600) 0.95 ( z) 0.05 z P( z 1.645) (iii) X 65 Z PZ (.467) 1 (.467) % (iv) Let Y represent the number of tins which overflow. Y B(1000, 0.006) EITHER: OR: Using the Poisson approximation to the binomial Mean = = 6. Y Poisson (6.) Using tables: P(Y > 10) 1 PY ( 10) Using the Normal approximation to the binomial, mean = , variance = Y N(6., ) Z MEI, 10/1/07 3/6

4 P(Y > 10.5) 1 PY ( 10.5 ) (i) We want P(9.5 < X < 30.5) X 9.5 Z X 30.5 Z P(-0.15 < Z < )= P( < Z < 0. 15) (0.15) (0.675) (ii) P(X < 9.5) ( 0.15 ) 1 (0.15 ) P(just one of the three gets fewer than 30) (iii)we want P(X > n) = 0. P(X < n) = 0. n 36 1 (0.) n so the lowest mark to obtain a grade A should be 43. (iv) We need P(X > 44.5) = 0.5 P(X < 44.5) = (0.75) The required mean mark should be (i) The number of arrivals are random and independent, and occur at a uniform rate. (ii) P(exactly arrivals) e MEI, 10/1/07 4/6

5 (iii)let X be the number of arrivals in a five-minute interval X Poisson (3.9) P(X 4) = 1 P(X 3) = = (iv) Let Y be the number of arrivals in 60 minutes Y Poisson (46.) Using the normal approximation to the Poisson distribution Y N(46., 46.) We want P(X > 40.5) X 40.5 Z P(X > 40.5) 1 ( 0.909) (0.909) 0.15 (v) Y N(60, 60 ) We want P(X > 40.5) = 0.0 P(X < 40.5) = (0.9) so (vi) Let u u u u.054 u u.054u u u (i) (A) 5 X 5 Z P(Z > ) = 1 P(Z < ) = = 0.53 MEI, 10/1/07 5/6

6 (B) 15 X 15 Z P(Z < ) = 1 P(Z < ) = = P(15 < X < 5) = = 0.67 (ii) P(X < T) = 0.95 T 1 (0.95) T (iii) From (i) P(completes puzzle) = P(completes puzzle on 0 days) P(completes puzzle on 1 day) P(completes puzzle at least twice) (iv) P(Y < 5) = (0.99) MEI, 10/1/07 6/6