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Exemplar 91586 2013 Question One • The weight of individual guavas is normally distributed with a mean of 215 g with a standard deviation of 13.2 g. A canning factory accepts guava for processing in three categories: less than 200 g, at least 200 g but less than 220 g, and 220 g and over. • What proportion of guavas will weigh either less than 200 g or 220 g and over? What proportion of guavas will weigh either less than 200 g or 220 g and over? 200 - 215 = 1.136... 13.2 P(X < 200) = 0.1279 z1 = 200 - 220 13.2 P(X > 220) = 0.35242 z2 = 200 215 220 The proportion of guavas that will weigh either less than 200 g or 220 g and over is 0.4803 • Calculate the probability that a guava selected at random is over 205 g, given that it is under 220 g in weight. Calculate the probability that a guava selected at random is over 205 g, given that it is under 220 g in weight. 205 - 215 13.2 P(X < 205) = 0.22435 z1 = P(X < 220) = 1- 0.35242 = 0.64758 P(205 < X < 220) = 0.42323 205 215 220 The probability that a guava selected at random is over 205 g, given that it is under 220 g in weight= 0.42323 = 0.6536 0.64758 A grower has a contract to supply the factory with 5 000 pineapples. The pineapples are normally distributed with a mean weight of 1.3 kg and a standard deviation of 0.234 kg. The canning factory pays $2.70 per kilogram of pineapple. The grower is hoping to earn $17 500 from his contract with the factory. By considering the variation of the pineapple weights, determine if the grower’s expectations for income from this contract are reasonable. Support your answer with calculations. A grower has a contract to supply the factory with 5 000 pineapples. The pineapples are normally distributed with a mean weight of 1.3 kg and a standard deviation of 0.234 kg. E ( X1 + X2 +...+ X5000 ) = 1.3+1.3+...+1.3 = 6500kg Var ( X1 + X2 +...+ X5000 ) = 0.234 2 + 0.234 2 +...+ 0.234 2 = 273.78 s T = 273.78 = 16.546.. The canning factory pays $2.70 per kilogram of pineapple. The grower is hoping to earn $17 500 from his contract with the factory. E ( 2.70T ) = 2.7 ´ 6500 = $17550 Var ( 2.7T ) = 2.7 2 ´ 273.78 s T = 1995.8562 = $44.68 By considering the variation of the pineapple weights, determine if the grower’s expectations for income from this contract are reasonable. Support your answer with calculations. E ( 2.70T ) = $17550, s T » $45 Since the expected value for the income if $17 550, with a standard deviation of only approximately $45, it is reasonable that the grower could gain an income close enough to $17 500 (the coefficient of variation is only 0.3% (45/17550). Alternate Answer 17500 -17550 z= 44.7 P ( X > 17500 ) = 0.8683 17500 17550 Alternate Answer 95% of income is expected to lie within 2 standard deviations from the mean i.e. Between $17460 and $17640 So it is reasonable to expect an income of $17 500 as this amount lies within the range for the central 95% of incomes. Question Two • The observations of the number of people who queue jumped (pushed ahead of others waiting to be served) at the express checkout at the local supermarket in a large number of 15-minute intervals produced the following graph. Calculate the expected number of queue jumpers in any given 15-minute interval. Calculate the expected number of queue jumpers in any given 15-minute interval. E ( X ) = 0 ´ 0.3+1´ 0.35 + 2 ´ 0.2 + 3´ 0.15 =1.2 Do not round to ‘1’ because the question says “expected number” E ( X ) = 0 ´ 0.3+1´ 0.35 + 2 ´ 0.2 + 3´ 0.15 =1.2 The store manager makes a statement to staff that he wants the store to be known as ‘queue–jump free’. This is defined by the manager as no more than one queue jumper per half an hour. Using your answer to part (a) and an appropriate distribution to model this situation, calculate the probability of there being no more than one queue jumper in any given half-hour period. In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Highlighting The store manager makes a statement to staff that he wants the store to be known as ‘queue–jump free’. This is defined by the manager as no more than one queue jumper per half an hour. Using your answer to part (a) and an appropriate distribution to model this situation, calculate the probability of there being no more than one queue jumper in any given half-hour period. In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Question states a ‘rate’ – Poisson distribution no more than one queue jumper per half an hour. l = 2.4 / ( 30 mins) Question states a ‘rate’ – Poisson distribution l = 2.4 / ( 30 mins) In your answer, you should justify your choice of distribution, identify the parameters of this distribution, Applying this distribution because: • discrete (queue jumpers) within a continuous interval (time) • it cannot occur simultaneously (only one queue jumper at a time) • queue jumping is a random event with no pattern to it • for a small interval (eg half an hour) the mean number of occurrences (queue jumpers) is proportional to the size of the interval. l = 2.4 / ( 30 mins) and state any assumptions you make Assumption is that each queue jumper is independent of other queue jumpers. Question states a ‘rate’ – Poisson distribution Using your answer to part (a) and an appropriate distribution to model this situation, calculate the probability of there being no more than one queue jumper in any given half-hour period. l = 2.4 / ( 30 mins) P(X ≤ 1) = 0.3084 • During the four weeks after the store manager made this statement, it was observed that the proportion of 10-minute periods with no queue jumpers was 90%. • • Can the store manager now claim that the store is ‘queue-jump free’? You should provide statistical evidence to support your answer. • During the four weeks after the store manager made this statement, it was observed that the proportion of 10-minute periods with no queue jumpers was 90%. P ( X = 0 ) = 0.9 = e- l ln 0.9 = -l l = 0.105 / (10 min ) l = 0.316 “no more than one queue jumper per half an hour” Can the store manager now claim that the store is ‘queue-jump free’? You should provide statistical evidence to support your answer. l = 0.316 P ( X £ 1) = 0.9594 The probability of there being more than one queue jumper per half an hour is less than 5%, so this could be evidence of the store being ‘queue-jump free’. Question Three Plants generally produce several flowers each season that need to be pollinated in order for fruit to be produced. A particular fruit plant produces one flower in one season per year. The flower must be pollinated in order for a single fruit to be produced. This plant type can be pollinated by bees carrying pollen from flower to flower between plants of the same type. Bee pollination has a 67% successful pollination rate. • Using an appropriate distribution, calculate the probability that one plant produces at least three fruit over four years (seasons), if bee pollination is used. In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Binomial Distribution In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Applying this distribution because: • fixed number of trials (four seasons/years) • fixed probability (67% success rate for pollination) • two outcomes (pollinated, not pollinated) • independence of events (a plant being pollinated or not does not affect chances of the same plant being pollinated or not in another season/year). Binomial Distribution In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Parameters: n = 4, π = 0.67 Binomial Distribution In your answer, you should justify your choice of distribution, identify the parameters of this distribution, and state any assumptions you make. Assumption is that the bees visit all the plants. calculate the probability that one plant produces at least three fruit over four years (seasons), if bee pollination is used. • P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.4015 = 0.5985 This plant type can also be pollinated by hand, with a person transferring pollen from flower to flower between plants of the same type. Hand pollination has a 91% successful pollination rate. A home grower has 12 plants of this plant type. Calculate the probability that for both of two years (seasons), only one of the 12 plants does not produce fruit if hand pollination is used. State any assumptions you make. Binomial distribution n = 12, p = 0.91 P(X = 11) = 0.3827 P(11 plants produce fruit for both seasons/years) = 0.38272 = 0.1465 Independence Assuming whether a plant is pollinated or not in one season does not affect the chances of the same plant being pollinated or not in another season/year, and assuming whether a plant is pollinated or not in one season/year does not affect the chances of another plant being pollinated or not in the same season/year. A commercial grower of this particular fruit plant would like to use a combination of both bee pollination and hand pollination, as bee pollination is much cheaper than hand pollination. The grower has 150 plants, of which they are contracted to provide 100 pollinated flowers to a local plant shop. The grower is considering using bee pollination for 90 of the plants, and hand pollination for 60 of the plants. The graph below shows the distributions of number of plants pollinated using each method. Comment on how likely it is that the grower will fulfil his contract of providing 100 pollinated flowers using this combination of pollination methods. Support your answer with appropriate calculations and assumptions. • Using parameters given earlier Bee: m = np = 90 ´ 0.67 = 60.3 s = 90 ´ 0.67 ´ 0.33 = 4.46 Hand: m = np = 60 ´ 0.91 = 54.6 s = 60 ´ 0.91´ 0.09 = 2.22 • Parameters for Normal distribution E ( B + H ) = 60.3+ 54.6 = 114.9 s B+H = 4.46 + 2.22 = 4.98 2 2 Using the Normal distribution E ( B + H ) = 60.3+ 54.6 = 114.9 s B+H = 4.46 2 + 2.22 2 = 4.98 P (T ³ 100 ) = P (T > 99.5 continutity correction ) = 0.99901 Using 2 standard deviations (10) from the mean (115) Range for central 95%, 105 < T < 124 So it is very likely that the grower will fulfil his contract using this combination of methods. • You could have used estimates from the graph but this would have been more difficult. Note the marking per question