To solve this problem, we will use the standard normal distribution (Z-distribution) since the thermometer readings are normally distributed with a mean ($\mu$) of $0^\circ\text{C}$ and a standard deviation ($\sigma$) of $1^\circ\text{C}$.

Step-by-Step Solution:

1. Define the Problem:

   • Mean ($\mu$) = $0^\circ\text{C}$
   • Standard Deviation ($\sigma$) = $1^\circ\text{C}$
   • Find the probability $P(0 \leq X \leq 0.94)$, where $X$ is the reading.

2. Convert to Z-scores: The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ For $X = 0$: $Z_1 = \frac{0 - 0}{1} = 0$ For $X = 0.94$: $Z_2 = \frac{0.94 - 0}{1} = 0.94$

3. Find the Corresponding Probabilities: Using a standard normal table or a calculator:

   • For $Z = 0$, the cumulative probability is: $P(Z \leq 0) = 0.5$
   • For $Z = 0.94$, the cumulative probability is approximately: $P(Z \leq 0.94) = 0.8264$

4. Calculate the Desired Probability: The probability of obtaining a reading between $0^\circ\text{C}$ and $0.94^\circ\text{C}$ is: $P(0 \leq X \leq 0.94) = P(Z \leq 0.94) - P(Z \leq 0)$ Substituting the values: $P(0 \leq X \leq 0.94) = 0.8264 - 0.5 = 0.3264$

Final Answer:

The probability of obtaining a reading between $0^\circ\text{C}$ and $0.94^\circ\text{C}$ is 0.3264 (or 32.64%).

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Would you like further clarification or additional examples?

Here are 5 related questions to expand your understanding:

1. What is the probability of obtaining a reading less than $-0.5^\circ\text{C}$?
2. How would the probability change if the standard deviation was 0.5°C instead of 1°C?
3. What is the probability of obtaining a reading greater than $1.5^\circ\text{C}$?
4. How can we calculate probabilities for readings in a non-standard normal distribution?
5. How do cumulative distribution functions (CDFs) help in solving such problems?

Tip: Always check if the distribution is standard normal or requires a Z-score transformation!