Moderate -0.3 This is a straightforward normal distribution problem requiring standardization and use of percentage points tables. Students must find σ from the given 5% condition, then apply it to find the second percentage. While it involves two steps, both are routine applications of the same technique with no conceptual challenges beyond standard S2 material.
The mass, in kilograms, of a packet of flour is a normally distributed random variable with mean 1.03 and variance \(\sigma^2\). Given that 5% of packets have mass less than 1.00 kg, find the percentage of packets with mass greater than 1.05 kg. [6]
1.645 seen anywhere, allow \(-1.645\), can be implied
\(= 1 - 0.8635 = \mathbf{0.1365}\) or \(13.6(5)\%\)
*M1
Solve to find \(\sigma\), or eliminate \(\sigma\), dependent on first M1
M1
Standardise with \(\mu = 1.03\), use \(\Phi\), answer \(< 0.5\), allow \(\sqrt{}\) errors
A1
Final answer in range \([0.1355, 0.137]\) or \([13.55\%, 13.7\%]\), must be from positive \(\sigma\), not from \(\sigma^2\) (6 marks)
0.1333 from \(\sigma = 0.018\) is 5+A0
$\frac{1.03 - 1.00}{\sigma} = 1.645$ | M1dep* | Standardise and equate to $\Phi^{-1}$, allow wrong sign, $\sigma^2$, $1 -$ , cc etc
$[\sigma = 0.0182... = ^{6}/_{320}]$ | A1 | All correct apart possibly from value of $\Phi^{-1}$
$1 - \Phi\left(\frac{1.05 - 1.03}{\sigma}\right) = 1 - \Phi(1.0966)$ | B1 | 1.645 seen anywhere, allow $-1.645$, can be implied
$= 1 - 0.8635 = \mathbf{0.1365}$ or $13.6(5)\%$ | *M1 | Solve to find $\sigma$, or eliminate $\sigma$, dependent on first M1
| M1 | Standardise with $\mu = 1.03$, use $\Phi$, answer $< 0.5$, allow $\sqrt{}$ errors
| A1 | Final answer in range $[0.1355, 0.137]$ or $[13.55\%, 13.7\%]$, must be from positive $\sigma$, not from $\sigma^2$ (6 marks) | 0.1333 from $\sigma = 0.018$ is 5+A0
The mass, in kilograms, of a packet of flour is a normally distributed random variable with mean 1.03 and variance $\sigma^2$. Given that 5% of packets have mass less than 1.00 kg, find the percentage of packets with mass greater than 1.05 kg. [6]
\hfill \mbox{\textit{OCR S2 2016 Q2 [6]}}