Question: Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P98, the 98-percentile. This is the temperature reading separating the bottom 98% from the top 2%. P98 = °C
1. **State the problem:** We need to find the 98th percentile ($P_{98}$) of a normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$. This percentile separates the lowest 98% of readings from the highest 2%.
2. **Recall the formula:** For a normal distribution, the percentile $P_p$ corresponding to a cumulative probability $p$ is given by
$$P_p = \mu + z_p \sigma$$
where $z_p$ is the z-score such that $P(Z \leq z_p) = p$ for the standard normal distribution $Z \sim N(0,1)$.
3. **Find the z-score for 98th percentile:** Using standard normal distribution tables or a calculator, the z-score for $p=0.98$ is approximately
$$z_{0.98} = 2.054$$
4. **Calculate $P_{98}$:** Substitute the values into the formula:
$$P_{98} = 0 + 2.054 \times 1 = 2.054$$
5. **Interpretation:** The temperature reading that separates the bottom 98% from the top 2% of thermometer readings is approximately $2.054^\circ C$.