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%.

Question

P70 =
°C

(Round answer to three decimal places)

Answer 1

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: \[ P70 = 0.524°C \]