Moderate -0.8 This is a straightforward application of normal distribution requiring students to find P(24.7 < X < 25.7), standardize to z-scores, use tables, and multiply by 500. It's routine with clear parameters and a standard method, making it easier than average but not trivial since it requires correct interpretation of 'within 0.5 cm' and accurate table reading.
2 A company produces a particular type of metal rod. The lengths of these rods are normally distributed with mean 25.2 cm and standard deviation 0.4 cm . A random sample of 500 of these rods is chosen.
How many rods in this sample would you expect to have a length that is within 0.5 cm of the mean length?
\(P\!\left(\frac{25.2-(25.5+0.50)}{0.4} < z < \frac{25.2-(25.2-0.50)}{0.4}\right)\) \(= P\!\left(-\frac{0.5}{0.4} < z < \frac{0.5}{0.4}\right)\)
M1
Use of \(\pm\) standardisation formula once; no continuity correction, \(\sigma^2\), \(\sqrt{\sigma}\)
\([= 2\Phi(1.25)-1]\)
A1
For AWRT \(0.8944\) SOI
\(= 2 \times 0.8944 - 1\)
M1
Appropriate area \(2\Phi - 1\) OE, from final process, must be probability
\(0.7888\)
A1
Accept AWRT \(0.789\)
Number of rods \(= 0.7888 \times 500 = 394\) or \(395\)
B1FT
Correct or FT *their* 4SF (or better) probability, final answer must be positive integer, not \(394.0\) or \(395.0\), no approximation/rounding stated, only 1 answer
5
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(\frac{25.2-(25.5+0.50)}{0.4} < z < \frac{25.2-(25.2-0.50)}{0.4}\right)$ $= P\!\left(-\frac{0.5}{0.4} < z < \frac{0.5}{0.4}\right)$ | M1 | Use of $\pm$ standardisation formula once; no continuity correction, $\sigma^2$, $\sqrt{\sigma}$ |
| $[= 2\Phi(1.25)-1]$ | A1 | For AWRT $0.8944$ SOI |
| $= 2 \times 0.8944 - 1$ | M1 | Appropriate area $2\Phi - 1$ OE, from final process, must be probability |
| $0.7888$ | A1 | Accept AWRT $0.789$ |
| Number of rods $= 0.7888 \times 500 = 394$ or $395$ | B1FT | Correct or FT *their* 4SF (or better) probability, final answer must be positive integer, not $394.0$ or $395.0$, no approximation/rounding stated, only 1 answer |
| | **5** | |
2 A company produces a particular type of metal rod. The lengths of these rods are normally distributed with mean 25.2 cm and standard deviation 0.4 cm . A random sample of 500 of these rods is chosen.
How many rods in this sample would you expect to have a length that is within 0.5 cm of the mean length?\\
\hfill \mbox{\textit{CAIE S1 2021 Q2 [5]}}