Thornleigh Salesian College, Bolton
Stats Easter School: 5. Estimation
Q1.The volume of bleach in a 5-litre bottle may be modelled by a random variable with a
standard deviation of 75 millilitres.
The volume, in litres, of bleach in each of a random sample of 36 such bottles was
measured. The 36 measurements resulted in a total volume of 181.80 litres and exactly 8
bottles contained less than 5 litres.
(a)
Construct a 98% confidence interval for the mean volume of bleach in a 5-litre
bottle.
(5)
(b)
It is claimed that the mean volume of bleach in a 5-litre bottle exceeds 5 litres and
also that fewer than 10 per cent of such bottles contain less than 5 litres.
Comment, with numerical justification, on each of these two claims.
(3)
(c)
State, with justification, whether you made use of the Central Limit Theorem in
constructing the confidence interval in part (a).
(1)
(Total 9 marks)
Q2.The weight, W grams, of a bag of sugar may be assumed to be normally distributed with a
mean of 1018 and a standard deviation of 10.
(a)
Determine the probability that:
(i)
W is less than 1025;
(2)
(ii)
W is greater than 1015 but less than 1030.
(3)
(b)
A box contains 24 such bags of sugar. Assuming that these 24 bags may be
regarded as a random sample, determine the probability that their mean weight
exceeds 1015 grams.
(4)
(Total 9 marks)
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Thornleigh Salesian College, Bolton
Q3.A random sample of 50 full-time university employees was selected as part of a higher
education salary survey.
The annual salary in thousands of pounds, x, of each employee was recorded, with the
following summarised results.
∑ x = 2290.0
and
∑ (x − x̅)2 = 28 225.50
Also recorded was the fact that 6 of the 50 salaries exceeded £60 000.
(a)
(i)
Calculate values for x̅ and s, where s2 denotes the unbiased estimate of σ 2.
(3)
(ii)
Hence show why the annual salary, X, of a full-time university employee is
unlikely to be normally distributed. Give numerical support for your answer.
(2)
(b)
(i)
Indicate why the mean annual salary, X̅, of a random sample of 50 full-time
university employees may be assumed to be normally distributed.
(2)
(ii)
Hence construct a 99% confidence interval for the mean annual salary of
full-time university employees.
(4)
(c)
It is claimed that the annual salaries of full-time university employees have an
average which exceeds £55 000 and that more than 25% of such salaries exceed
£60 000.
Comment on each of these two claims.
(3)
(Total 14 marks)
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Thornleigh Salesian College, Bolton
M1.(a)
x̅ =
5.05 or 5050
CAO
B1
98% (0.98) ⇒ z = 2.32 to 2.33
AWFW
(2.3263)
B1
CI for μ is
Used with z (2.05 to 2.58), x̅ (5.05, 5050 or 181.8),
σ (0.0075, 0.075, 0.75, 7.5 or 75)
and ÷
with n > 1
M1
Thus 5.05 ≠ 2.3263 ×
z (2.05 to 2.06 or 2.32 to 2.33 or 2.57 to 2.58),
x̅ (5.05) & σ (0.075) or x̅ (5050) & σ (75)
and ÷
A1
Hence 5.05 ± 0.03 or 5050 ± 30
OR
(5.02, 5.08) or (5020, 5080)
CAO / AWRT
Dependent on previous A1 so can be scored with
z ≠ 2.32 to 2.33
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Thornleigh Salesian College, Bolton
Ignore (absence of) quoted units
AWRT to 3sf accuracy
Adep1
5
Note
(b)
Use of t(2.43 to 2.72) ⇒ B1 B0 M1 A0 A0 max
Clear correct comparison of 5 or 5000 with LCL or CI
so
agree with (first) claim (about mean)
Dependent on Adep1 in (a)
Must use consistent units
Adep1
(8 / 36 or 0.22 or 22%) v (1 / 10 or 0.10 or 10%)
or 8 v 3.6 (3 to 4)
Mention of a value on LHS and a value on RHS
B1
so
8 / 36 OE >/≠ 1 / 10 OE
so
disagree with (second) claim (about individuals)
Dependent on B1
Explicit comparison of values and correct conclusion
Bdep1
3
Notes 1 It / (claimed) mean / (claimed) value < LCL / CI ⇒ Adep0
Must indicate 5 or 5000
2 98% have (mean) weights between CLs so … ⇒
Adep0
3 Any reference to CI for second claim ⇒ B0 Bdep0
Claim refers to individual bottles
(c)
Yes because volumes / bleach / litres / bottles / (parent) population
are not (stated as) normally distributed
OE; but do not accept ‘data’ or ‘sample’ or ‘it’
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Thornleigh Salesian College, Bolton
Reference to sample size only ⇒ B0
(eg n > 25 or n > 30)
B1
1
[9]
M2.(a)
Weight, W ̴ N (1018, 102)
(i)
P(W < 1025) =
Standardising (1024.5, 1025 or 1025.5) with 1018
and (√10, 10 or 102) and / or (1018 – x)
May be gained in (a)(i) or (a)(ii)
M1
P(Z < 0.7) = 0.758
AWRT (0.75804)
A1
2
(ii)
P(1015 < W < 1030)
= P (W < 1030) – P(W < 1015)
= P(Z < 1.2) – P(Z < –0.3)
Difference of two probabilities
May be implied
M1
= 0.88493 – (1 – 0.61791)
Area change
May be implied by answer
M1
= 0.502 to 0.504
AWFW (0.50284)
A1
3
(b)
̴ Normal with μ = 1018 and
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Thornleigh Salesian College, Bolton
to 4.2
CAO/AWFW (4.16666)
B1
or
to 2.05
CAO/AWFW (2.04124)
Standardising 1015 with 1018 and c’s
; not σ
M1
= P (Z > –1.47) = 1 – P(Z <1.47)
Area change
May be implied by answer > 0.5
m1
= 0.927 to 0.932
AWFW (0.92918)
A1
4
[9]
M3.(a)
(i)
x̅ =
= 45.8 or 45800
CAO
B1
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Thornleigh Salesian College, Bolton
(s2 =)
or (s =)
Ignore notation
M1
s = 24(.0) or 24000 to 24001
AWRT / AWFW
(σ = 23.75942)
(24.00064)
SCs: (for no seen working)
M1 A1 for 24.0 or 24000 to 24001
M1 A0 for 24 or 23700 to 23800
A1
3
(ii)
x̅ − ns = (45.8 − n × 24.0)
SC: Accept quoted values of
(−4 to −1) (n = 2) or (−28.5 to −23.5) (n = 3)
(both AWFW)
Allow (45 to 47) and any multiple of
(23.5 to 24.5) which gives value < 0
Must clearly state the value of a numerical expression
M1
and negative salaries are impossible
OE; must be in context
Negative values impossible ⇒
A0
A1
2
Alternative solution 1
P(X 2) = P(Z < −1.91)
Standardising 0 using 45.8 & 24.0
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Thornleigh Salesian College, Bolton
(M1)
= 0.027 to 0.03
In addition to probability within range,
must state that negative salaries are impossible
(A1)
(2)
Alternative solution 2
P(X > 60 | N(45.8, 24.02)) = P(Z > 0.59)
Standardising 60 using 45.8 & 24.0
(M1)
= 0.27 to 0.28
In addition to probability within range,
must compare calculated value to 6 / 50 = 0.12 OE
(A1)
(2)
(b)
(i)
Large sample or n > 25 or 30 or n = 50
OE
B1
so CLT applies
Must indicate CLT; dependent on B1
Indication that other than sample mean
is normally distributed ⇒ Bdep0
Bdep1
2
(ii)
99% (0.99) ⇒ z = 2.57 to 2.58
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Thornleigh Salesian College, Bolton
AWFW
(2.5758)
B1
CI for μ is
x̅ ± z ×
Used with (x̅ & s) from (a)(i) and
z(1.64 to 2.58) & ÷
with n > 1
M1
Thus
45.8 ± 2.5758 ×
F on (x̅ & s) with ÷
&
z(1.64 to 1.65 or 2.32 to 2.33 or 2.57 to 2.58)
AF1
Hence 45.8 ± (8.7 to 8.8)
or
45800 ± (8700 to 8800)
OR
(37.(0) to 37.1, 54.5 to 54.6)
or
(37000 to 37100, 54500 to 54600)
CAO / AWFW
(8.74)
Ignore (absence of) quoted units
AWFW
A1
4
(c)
See supplementary sheet for additional illustrations
Clear correct comparison of 55 or 55000 with c’s UCL or CI
Accept 55000 compared with
c’s 54.5 to 54.6 (ie different units)
B1
(6/50 or 0.12 or 12%) </≠ 0.25 or 25%
OE; correct comparison mentioning both 12% and 25%
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Thornleigh Salesian College, Bolton
B1
Reject both / each of the two claims
Dependent on B1 B1
Bdep1
3
Additional comment illustrations
It / (claimed) mean / (claimed) value > UCL / CI
Must indicate 55 or 55000
(B0)
99% have (mean) weights between CLs so...
(B0)
Any comparison of 60 (£60 000) with UCL / IC
Value of 60 does not refer to mean
(B0)
P(X < 60 | N(45.8, 24.02)) = P(Z > 0.59)
= (0.27 to 0.28) < 6 / 50 = 0.12
Assumes salaries ~ N; cf (a)(ii)
(B0)
[14]
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Thornleigh Salesian College, Bolton
E1.The better students generally scored full marks in part (a) usually by working in litres.
However, far too many students provided answers that ranged from worthless to scoring
at most 3 marks. Errors seen in abundance were:
•
an incorrect evaluation of
•
the use of 5, 181.80 or even
; this despite the use of a calculator;
instead of 5.05;
•
the use of 75 or 0.75 instead of 0.075;
•
an incorrect z–value;
•
the omission of the divisor of
.
Many of these errors resulted in absurd confidence intervals that students appeared to
simply accept as correct. The one mark for answering the first claim in part (b) was only
available to those better students who had obtained a correct confidence interval in part
(a)(ii). Even so, many such students made no comparison using the stated mean of 5
litres but merely commented that their calculated mean of 5.05 was greater than 5. This
was very disappointing particularly given similar requests on previous papers. In
answering the second claim in part (b), many students again referred to their confidence
interval and so scored no marks. Of those that correctly evaluated
as 22%, far too
many were not sufficiently explicit in comparing this with 10% and so scored only 1 of the
2 marks available. It was extremely rare to see an answer worthy of the mark in part (c).
Most students made reference to sample size although some suggested that the question
had stated that volume was normally distributed! The minority who had some idea of what
was required were almost always not sufficiently precise or careful enough in their
answers. Thus it was common to read references to ‘the sample’, ‘the data’ or ‘it’ rather
than for example ‘the (parent) population’ or ‘volume (of bleach)’.
E2.Most candidates achieved the 5 marks available in part (a) but very few were able to make
significant progress in part (b). This was disappointing, particularly as the distribution of
the sample mean has been examined on several previous papers. Almost all candidates
knew how to standardise – without introducing an unnecessary continuity correction,
which was penalised – and so the majority completed part (a)(i) accurately. In part (a)(ii),
the majority of candidates realised that a difference of two areas was required, with many
obtaining the correct answer. Failure to apply the necessary area change for P (Z < − 0.3)
was by far the most common error. The majority of candidates simply worked with the
distribution of W, rather than
, and so scored no marks. Candidates who worked with
the distribution of
, and so obtained correctly P (Z > − 147), sometimes then failed to
make the correct area change and so lost 2 marks.
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Thornleigh Salesian College, Bolton
E3.Most candidates found much of this question very challenging. In part (a)(i), it was most
disappointing to see the number of candidates failing to score 3 marks. Whilst there was
some excuse for candidates dividing ∑ (x – x̅)2 by 50, rather than 49, there was really no
excuse for the many who did not know that x̅ =
. Frequent incorrect attempts
included trying to substitute ∑ x into ∑(x – x̅)2 or dividing ∑ x by a variety of
denominators. Some candidates were confused, either at the outset or later, by the units
involved with invalid multiplications of s2 by 1000 taking place at some point. In part (a)(ii),
the common incorrect answers were verbose about university salaries or simply that 24.0
was too large when compared to 45.8. Even those candidates who calculated, for
example (45.8 – 2 × 24.0), simply stated that a normal distribution or ‘it’, instead of
salaries, could not take negative values. Answers to part (b)(i) were slightly better with
candidates indicating a ‘large sample’ so ‘Central Limit Theorem applied’. Some
candidates lost both marks for ‘if sample is large’ whilst others lost a mark for suggesting
that other than the sample mean could be assumed to be normally distributed. There were
many partially correct answers to part (b)(ii) using follow-through answers from part (a)(i)
though some candidates lost further marks for an incorrect z-value. In answering part (c),
a majority of candidates attempted the comparison of 55 with their CI or UCL. However,
far too many then did the same with 60 rather than compare
= 12% with 25%, or
considered the standardising of 60 using a normal distribution. This perhaps suggested
that they had not read the question with sufficient care.
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