1.9 Practice – Age Problems
Ages (in years) of licensed drivers in a region are modeled as approximately normal with mean \(\mu=38\) and standard deviation \(\sigma=12\). Denote age by \(X\), so \(X \sim N(38,12^2)\).
Compute the following:
1) \(P(25 \le X \le 50)\)
2) \(P(X > 60)\)
3) The 90th percentile age (the value \(x\) such that \(P(X \le x)=0.90\))
4) In a random sample of \(n=200\) drivers, the expected number older than 60
Key idea: standardize to a z-score
Standardizing converts an \(x\)-value from the normal model into a standard normal value:
\[ z=\frac{x-\mu}{\sigma}. \]
After standardizing, probabilities are read from the standard normal cumulative distribution function \(\Phi(z)\), where \(\Phi(z)=P(Z \le z)\) and \(Z \sim N(0,1)\).
1) Probability between two ages: \(P(25 \le X \le 50)\)
- Standardize the endpoints: \[ z_{25}=\frac{25-38}{12}=\frac{-13}{12}\approx -1.0833, \qquad z_{50}=\frac{50-38}{12}=\frac{12}{12}=1.0000. \]
- Convert to a standard normal probability: \[ P(25 \le X \le 50)=P(z_{25}\le Z \le z_{50})=\Phi(1.0000)-\Phi(-1.0833). \]
- Use typical table/CDF values (rounded): \(\Phi(1.00)\approx 0.8413\) and \(\Phi(-1.0833)\approx 0.1391\). Then \[ P(25 \le X \le 50)\approx 0.8413-0.1391=0.7022. \]
Result: \(P(25 \le X \le 50)\approx 0.702\).
2) Probability above an age: \(P(X > 60)\)
- Standardize 60: \[ z_{60}=\frac{60-38}{12}=\frac{22}{12}\approx 1.8333. \]
- Right-tail probability: \[ P(X > 60)=P(Z > 1.8333)=1-\Phi(1.8333). \]
- Using a typical table/CDF value \(\Phi(1.8333)\approx 0.9666\): \[ P(X > 60)\approx 1-0.9666=0.0334. \]
Result: \(P(X > 60)\approx 0.033\).
3) 90th percentile age
The 90th percentile is the value \(x\) for which \(P(X \le x)=0.90\). On the standard normal scale, this corresponds to a z-value \(z_{0.90}\) such that \(\Phi(z_{0.90})=0.90\).
- Use the inverse-CDF (or a standard table): \(z_{0.90}\approx 1.2816\).
- Convert back to the original age scale: \[ x=\mu+z_{0.90}\cdot\sigma=38+1.2816\cdot 12. \]
- Compute: \[ x=38+15.3792=53.3792 \approx 53.4. \]
Result: the 90th percentile age is approximately \(53.4\) years.
4) Expected number older than 60 in \(n=200\)
If each selected driver is older than 60 with probability \(p=P(X > 60)\approx 0.0334\), then the expected count is
\[ E=\;n\cdot p=200\cdot 0.0334=6.68 \approx 7. \]
| Quantity | Key computation | Approximate value |
|---|---|---|
| \(P(25 \le X \le 50)\) | \(\Phi(1.0000)-\Phi(-1.0833)\) | \(0.702\) |
| \(P(X > 60)\) | \(1-\Phi(1.8333)\) | \(0.033\) |
| 90th percentile | \(38+1.2816\cdot 12\) | \(53.4\) years |
| Expected count in \(n=200\) | \(200\cdot 0.0334\) | \(\approx 7\) |