OCR S3 2013 June — Question 1 6 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2013
SessionJune
Marks6
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TopicLinear combinations of normal random variables
TypeSingle variable sum probability
DifficultyStandard +0.3 This is a straightforward application of the sum of independent normal random variables. Students need to recognize that the sum of 38 independent N(3.5, 0.9²) variables is N(38×3.5, 38×0.9²), then standardize and use tables. It's slightly above average difficulty due to the context interpretation and calculation with 38 variables, but requires no novel insight—just direct application of a standard S3 result.
Spec5.04b Linear combinations: of normal distributions
1 The blood-test procedure at a clinic is that a person arrives, takes a numbered ticket and waits for that number to be called. The waiting times between the numbers called have independent normal distributions with mean 3.5 minutes and standard deviation 0.9 minutes. My ticket is number 39 and as I take my ticket number 1 is being called, so that I have to wait for 38 numbers to be called. Find the probability that I will have to wait between 120 minutes and 140 minutes.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Total time \(T \sim N(\mu, \sigma^2)\)M1 Using \(\sum T_i \sim N\)
\(\mu = 38 \times 3.5\)A1 \(= 133\); 129.5 (from 37) or 136.5 (from 39) A0
\(\sigma^2 = 38 \times 0.9^2\)A1 \(= 30.78\); 29.97 (from 37) or 31.59 (from 39) A0
\(P(120 < T < 140)\)M1 M1 for standardising and combining. Allow even if spurious cc, or \(\sigma^2\) from \(38^2 \times 0.9^2\) used
\(= \Phi[(140-133)/\sigma] - \Phi[(120-133)/\sigma]\)
\(= 0.8966 - 0.0095\)A1 allow \(0.9724 - 0.0414\) (from 37) or \(0.7336 - 0.0017\) (from 39) A1ft
\(= 0.887\)A1 \(= 0.931\) or \(= 0.732\) A1ft
[6] 
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Total time $T \sim N(\mu, \sigma^2)$ | M1 | Using $\sum T_i \sim N$ |
| $\mu = 38 \times 3.5$ | A1 | $= 133$; 129.5 (from 37) or 136.5 (from 39) A0 |
| $\sigma^2 = 38 \times 0.9^2$ | A1 | $= 30.78$; 29.97 (from 37) or 31.59 (from 39) A0 |
| $P(120 < T < 140)$ | M1 | M1 for standardising and combining. Allow even if spurious cc, or $\sigma^2$ from $38^2 \times 0.9^2$ used |
| $= \Phi[(140-133)/\sigma] - \Phi[(120-133)/\sigma]$ | | |
| $= 0.8966 - 0.0095$ | A1 | allow $0.9724 - 0.0414$ (from 37) or $0.7336 - 0.0017$ (from 39) A1ft |
| $= 0.887$ | A1 | $= 0.931$ or $= 0.732$ A1ft |
| | **[6]** | |

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1 The blood-test procedure at a clinic is that a person arrives, takes a numbered ticket and waits for that number to be called. The waiting times between the numbers called have independent normal distributions with mean 3.5 minutes and standard deviation 0.9 minutes. My ticket is number 39 and as I take my ticket number 1 is being called, so that I have to wait for 38 numbers to be called. Find the probability that I will have to wait between 120 minutes and 140 minutes.

\hfill \mbox{\textit{OCR S3 2013 Q1 [6]}}
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