Edexcel S1 2002 November — Question 4 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2002
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeProbability calculation plus find unknown boundary
DifficultyModerate -0.3 This is a straightforward application of normal distribution with standard procedures: (a) uses inverse normal tables to find μ from a given percentage, (b) applies standardization and table lookup, (c) uses basic probability multiplication. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average for A-level.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

Strips of metal are cut to length \(L\) cm, where \(L \sim N(\mu, 0.5^2)\).
  1. Given that 2.5\% of the cut lengths exceed 50.98 cm, show that \(\mu = 50\). [5]
  2. Find \(P(49.25 < L < 50.75)\). [4]
Those strips with length either less than 49.25 cm or greater than 50.75 cm cannot be used. Two strips of metal are selected at random.
  1. Find the probability that both strips cannot be used. [2]

(a)
AnswerMarks
\(P(L > 50.98) = 0.025\)B1
\(P\left(Z > \frac{50.98 - \mu}{0.5}\right) = 0.025\)M1 A1
\(\frac{50.98 - \mu}{0.5} = 1.96\)M1 A1
\(\mu = 50\) (*)(5 marks)
(b)
AnswerMarks
\(P(49.25 < L < 50.75) = P\left(\frac{49.25 - 50}{0.5} < Z < \frac{50.75 - 50}{0.5}\right)\)M1
\(= P(-1.5 < Z < 1.5)\)A1
\(= 2\Phi(1.5) - 1\)M1
\(= 0.8664\)A1
(4 marks)
(c)
AnswerMarks
\(P(\text{Both}) = (1 - 0.8664)^2\)M1 A1
\(= 0.01784\ldots\)A1
(2 marks)
Total: 11 marks
## (a)
$P(L > 50.98) = 0.025$ | B1 |
$P\left(Z > \frac{50.98 - \mu}{0.5}\right) = 0.025$ | M1 A1 |
$\frac{50.98 - \mu}{0.5} = 1.96$ | M1 A1 |
$\mu = 50$ (*) | (5 marks) |

## (b)
$P(49.25 < L < 50.75) = P\left(\frac{49.25 - 50}{0.5} < Z < \frac{50.75 - 50}{0.5}\right)$ | M1 |
$= P(-1.5 < Z < 1.5)$ | A1 |
$= 2\Phi(1.5) - 1$ | M1 |
$= 0.8664$ | A1 |
| (4 marks) |

## (c)
$P(\text{Both}) = (1 - 0.8664)^2$ | M1 A1 |
$= 0.01784\ldots$ | A1 |
| (2 marks) |

**Total: 11 marks**

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Strips of metal are cut to length $L$ cm, where $L \sim N(\mu, 0.5^2)$.

\begin{enumerate}[label=(\alph*)]
\item Given that 2.5\% of the cut lengths exceed 50.98 cm, show that $\mu = 50$. [5]

\item Find $P(49.25 < L < 50.75)$. [4]
\end{enumerate}

Those strips with length either less than 49.25 cm or greater than 50.75 cm cannot be used.

Two strips of metal are selected at random.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the probability that both strips cannot be used. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2002 Q4 [11]}}