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 P70, the 70-percentile. This is the temperature reading separating the bottom 70% from the top 30%.
P70 =
°C
(Round answer to three decimal places)
P70 =
°C
(Round answer to three decimal places)
Answers
GPT-4o mini
To find the 70th percentile (P70) of a normal distribution with a mean (μ) of 0°C and a standard deviation (σ) of 1.00°C, we can use the z-score formula and standard normal distribution tables or calculators.
1. **Find the z-score corresponding to the 70th percentile**: This can be found using z-tables or a calculator. The z-score associated with the cumulative probability of 0.70 is approximately \( z \approx 0.524 \).
2. **Use the z-score formula** to find the corresponding temperature reading:
\[
P(X \leq x) = μ + z \cdot σ
\]
Here, \( μ = 0°C \) and \( σ = 1.00°C \).
3. **Substitute the values**:
\[
P70 = 0 + 0.524 \cdot 1.00 = 0.524
\]
Thus, the 70th percentile (P70) is approximately:
\[
P70 \approx 0.524°C
\]
Finally, rounding to three decimal places gives:
\[
P70 = 0.524°C
\]
1. **Find the z-score corresponding to the 70th percentile**: This can be found using z-tables or a calculator. The z-score associated with the cumulative probability of 0.70 is approximately \( z \approx 0.524 \).
2. **Use the z-score formula** to find the corresponding temperature reading:
\[
P(X \leq x) = μ + z \cdot σ
\]
Here, \( μ = 0°C \) and \( σ = 1.00°C \).
3. **Substitute the values**:
\[
P70 = 0 + 0.524 \cdot 1.00 = 0.524
\]
Thus, the 70th percentile (P70) is approximately:
\[
P70 \approx 0.524°C
\]
Finally, rounding to three decimal places gives:
\[
P70 = 0.524°C
\]