Standard +0.3 This is a straightforward application of the Central Limit Theorem to a binomial distribution. Students need to find the mean and variance of X, apply CLT to get the distribution of the sample mean, then perform a standard normal calculation. While it requires multiple steps, each is routine and the question clearly signals which theorem to use.
\(X\) is a random variable having the distribution \(\text{B}(12, \frac{1}{4})\). A random sample of 60 values of \(X\) is taken. Find the probability that the sample mean is less than 2.8. [5]
(because this is inherent in the context so it’s not a
condition)
Poisson probs, λ = 1.8. Allow 2, 6
included
Any λ. Allow end errors.
[Total: 7]
4 (i)
(ii)
Answer
Marks
(iii)
x= 8.4
1.3
8.4 ± z
15
z = 2.576
[7.54, 9.26]
No because pop normal
soX normally distr
8 within CI
Answer
Marks
Claim justified
B1
M1
B1
A1 [4]
B1
B1 [2]
B1√
B1√
Answer
Marks
[2]
Accept 2.574 to 2.579
or equiv. Accept 7.53. Accept 9.27
SR If ‘Yes’ or no conclusion, but 2
correct statements score B1
ft (i)
[Total: 8]
Answer
Marks
Guidance
Page 5
Mark Scheme: Teachers’ version
Syllabus
GCE AS/A LEVEL – May/June 2011
9709
72
Question 2:
2 | 9
X ~N(3, 4 )
60
2.8−3
(= –1.033)
9
4
60
Φ(“–1.033”) = 1 – Φ(“1.033”)
= 0.151 | B2
M1
M1
A1 [5] | B1 for N & µ = 3; (oe)
9/4 3
B1 for /60 or /80 or 0.0375 (oe)
(oe working with totals or proportions)
With or without c.c.
With cc of –1/ , Φ(–1.076) = 1 – Φ(1.076) =
120
0.141
[Total: 5]
3 (i)
(ii)
(iii) | Constant average rate of goals scored
Goals random
Goals indep
–1.8(1.83 +1.84 +1.85
e )
3! 4! 5!
= 0.259
–1.8
1 – e
–1.8 10
(1 – e )
= 0.164 | B1
B1 [2]
M1
A1 [2]
M1
M1
A1 [3] | Any two given in context
(SR score B1 for any two not in context)
Not Goals scored singly
(because this is inherent in the context so it’s not a
condition)
Poisson probs, λ = 1.8. Allow 2, 6
included
Any λ. Allow end errors.
[Total: 7]
4 (i)
(ii)
(iii) | x= 8.4
1.3
8.4 ± z
15
z = 2.576
[7.54, 9.26]
No because pop normal
soX normally distr
8 within CI
Claim justified | B1
M1
B1
A1 [4]
B1
B1 [2]
B1√
B1√
[2] | Accept 2.574 to 2.579
or equiv. Accept 7.53. Accept 9.27
SR If ‘Yes’ or no conclusion, but 2
correct statements score B1
ft (i)
[Total: 8]
Page 5 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE AS/A LEVEL – May/June 2011 | 9709 | 72
$X$ is a random variable having the distribution $\text{B}(12, \frac{1}{4})$. A random sample of 60 values of $X$ is taken. Find the probability that the sample mean is less than 2.8. [5]
\hfill \mbox{\textit{CAIE S2 2011 Q2 [5]}}