Exemplar - Mags Maths

Report
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

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