Moderate -0.8 This is a straightforward inverse normal distribution problem requiring a single standardization and rearrangement. Students need only look up the z-value for probability 0.242, then solve μ = 22 - 5z. It's simpler than average A-level questions as it involves just one calculation step with no problem-solving insight required.
1 The random variable \(H\) has the distribution \(\mathrm { N } \left( \mu , 5 ^ { 2 } \right)\). It is given that \(\mathrm { P } ( H < 22 ) = 0.242\). Find the value of \(\mu\).
Standardise with \(\Phi^{-1}\), allow +, "1 −" errors, cc, \(\sqrt{5}\) or \(5^2\)
\(\mu = 25.8\)
A1
Correct equation including signs, no cc, can be wrong \(\Phi^{-1}\)
B1
0.7 correct to 3 SF, can be +
A1
Answer 25.5 correct to 3 SF
$\frac{22-\mu}{5} = -\Phi^{-1}(0.242)$ | M1 | Standardise with $\Phi^{-1}$, allow +, "1 −" errors, cc, $\sqrt{5}$ or $5^2$
$\mu = 25.8$ | A1 | Correct equation including signs, no cc, can be wrong $\Phi^{-1}$
| B1 | 0.7 correct to 3 SF, can be +
| A1 | Answer 25.5 correct to 3 SF
1 The random variable $H$ has the distribution $\mathrm { N } \left( \mu , 5 ^ { 2 } \right)$. It is given that $\mathrm { P } ( H < 22 ) = 0.242$. Find the value of $\mu$.
\hfill \mbox{\textit{OCR S2 2007 Q1 [4]}}