OCR S2 2005 June — Question 2 4 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2005
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind standard deviation from probability
DifficultyModerate -0.3 This is a straightforward inverse normal distribution problem requiring a single lookup of z = -0.674 from tables (since P(Z > z) = 0.75) and solving (20 - 25)/σ = -0.674. It's slightly easier than average because it's a direct one-step application with no complications, though it does require understanding of standardization.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
2 A continuous random variable has a normal distribution with mean 25.0 and standard deviation \(\sigma\). The probability that any one observation of the random variable is greater than 20,0 is 0.75 . Find the value of \(\sigma\).
AnswerMarks Guidance
\(20 - 25 = \Phi^{-1}(0.25) = -0.674\)M1 Standardise and equate to \(\Phi^{-1}\) [not .7754 or .5987]
\(\sigma = 0.674 = 7.42\)B1
M1\(z\) in range [-0.675, -0.674], allow \(\pm\)
A1 4\((\pm) 5 + z\)-value [not Φ(2) or 0.75]
Answer in range [7.41, 7.42], no sign digits [SR: \(\sigma^2\): M1B1M0A0; cc: M1B1M1A0] 
$20 - 25 = \Phi^{-1}(0.25) = -0.674$ | M1 | Standardise and equate to $\Phi^{-1}$ [not .7754 or .5987]
$\sigma = 0.674 = 7.42$ | B1 |
| M1 | $z$ in range [-0.675, -0.674], allow $\pm$
| A1 4 | $(\pm) 5 + z$-value [not Φ(2) or 0.75]
| | Answer in range [7.41, 7.42], no sign digits [SR: $\sigma^2$: M1B1M0A0; cc: M1B1M1A0]
2 A continuous random variable has a normal distribution with mean 25.0 and standard deviation $\sigma$. The probability that any one observation of the random variable is greater than 20,0 is 0.75 . Find the value of $\sigma$.

\hfill \mbox{\textit{OCR S2 2005 Q2 [4]}}
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