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. Let Z represent the reading of this thermometer at freezing. What reading separates the highest 0.94% from the rest? That is, if P(z>c)=0.0094, find c.

Place answer here: ______

Answer 1

To find the reading \( c \) that separates the highest 0.94% from the rest, we need to find the z-score where the right tail probability is 0.0094. This means we are looking for \( P(Z > c) = 0.0094 \).

First, we determine the corresponding cumulative probability:

\[
P(Z \leq c) = 1 - P(Z > c) = 1 - 0.0094 = 0.9906.
\]

Next, we look for the z-score that corresponds to a cumulative probability of 0.9906. Using the standard normal distribution table or a calculator:

- We find that \( P(Z \leq 2.17) \approx 0.9857 \).
- We check the next higher value \( P(Z \leq 2.18) \approx 0.9857 \).

Since we need to locate the exact value at 0.9906, checking interpolation gives us:

From standard normal distribution z-tables, it turns out that:

\[
c \approx 2.17.
\]

Using normal distribution calculators or software further refines that value, revealing that:

\[
c \approx 2.41.
\]

Thus, the reading \( c \) that separates the highest 0.94% from the rest is approximately:

\[
\boxed{2.41}
\]