We are given that the thermometer readings follow a normal distribution with:

• Mean: $\mu = 0$°C
• Standard deviation: $\sigma = 1.00$°C

We need to find the probability that a randomly selected thermometer gives a reading between -1.244°C and -0.374°C.

Step 1: Convert Raw Scores to Standard Scores (Z-Scores)

The standard normal variable $Z$ is given by:

$Z = \frac{X - \mu}{\sigma}$

For $X_1 = -1.244$:

$Z_1 = \frac{-1.244 - 0}{1} = -1.244$

For $X_2 = -0.374$:

$Z_2 = \frac{-0.374 - 0}{1} = -0.374$

Step 2: Find the Corresponding Probabilities

Using the standard normal table (or a calculator):

$P(Z \leq -1.244) \approx 0.1079$

$P(Z \leq -0.374) \approx 0.3544$

Step 3: Compute the Probability Between the Two Z-Scores

$P(-1.244 \leq Z \leq -0.374) = P(Z \leq -0.374) - P(Z \leq -1.244)$

$= 0.3544 - 0.1079$

$= 0.2465$

Final Answer:

The probability of obtaining a thermometer reading between -1.244°C and -0.374°C is 0.2465 (or 24.65%).

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Would you like me to explain how to use the Z-table or check with a different method?

Related Questions:

1. What is the probability of a reading above 1.5°C?
2. How do you calculate probabilities for any normal distribution?
3. What is the probability of getting a reading below -2°C?
4. How does changing the standard deviation affect probabilities?
5. What is the Z-score for a reading of 2.5°C?

Tip:

If you don't have a Z-table handy, you can use an online normal probability calculator or statistical functions in software like Excel (=NORM.DIST(x, mean, stdev, TRUE)).