Easy -1.2 This is a straightforward application of the coding formula for means: if Σ(x-10)=27, then Σx - 10n = 27, and since mean = Σx/n = 11.5, we get 11.5n - 10n = 27, giving n=18. Requires only basic algebraic manipulation of a standard coded sum formula, making it easier than average.
1 Andy counts the number of emails, \(x\), he receives each day and notes that, over a period of \(n\) days, \(\Sigma ( x - 10 ) = 27\) and the mean number of emails is 11.5 . Find the value of \(n\).
Expanding brackets and forming a three term equation involving 27 and at least one term in \(n\), without \(x\)
\(10n\) or \(11.5n\) seen in expression without \(x\)
M1
(\(1.5n = 27\) implies M2)
\(n = 18\)
A1)
OR:
Answer
Marks
Guidance
\(11.5 = \frac{27}{n} + 10\)
(M1)
Dividing coded sum by \(n\) and forming a three term equation involving 11.5 and at least one term in \(n\), without \(x\)
\(27/n\) seen in expression without \(x\)
M1
(\(1.5 = \frac{27}{n}\) implies M2)
\(n = 18\)
A1)
Total: 3 marks
## Question 1:
**EITHER:**
$(\Sigma x =)\ 11.5n = 27 + 10n$ | **(M1)** | Expanding brackets and forming a three term equation involving 27 and at least one term in $n$, without $x$
$10n$ or $11.5n$ seen in expression without $x$ | **M1** | ($1.5n = 27$ implies **M2**)
$n = 18$ | **A1)** |
**OR:**
$11.5 = \frac{27}{n} + 10$ | **(M1)** | Dividing coded sum by $n$ and forming a three term equation involving 11.5 and at least one term in $n$, without $x$
$27/n$ seen in expression without $x$ | **M1** | ($1.5 = \frac{27}{n}$ implies **M2**)
$n = 18$ | **A1)** |
**Total: 3 marks**
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1 Andy counts the number of emails, $x$, he receives each day and notes that, over a period of $n$ days, $\Sigma ( x - 10 ) = 27$ and the mean number of emails is 11.5 . Find the value of $n$.\\
\hfill \mbox{\textit{CAIE S1 2017 Q1 [3]}}