Assume that the readings on the thermometers are normally distributed with a mean of 0∘ and a standard deviation of 1.00∘C. A thermometer is randomly selected and tested. Find the probability of each reading in degrees.
(a)   Between 0 and 1.23:
(b)   Between −2.65 and 0:
(c)   Between −0.52 and 1.51:
(d)   Less than −2.62:
(e)   Greater than 2.89:

To solve this, we need to use the properties of the standard normal distribution, where the mean \( \mu = 0 \) and the standard deviation \( \sigma = 1 \). We can use standard normal tables or a calculator to find the cumulative probabilities associated with specific \( z \)-scores.

Here's the breakdown for each part:

### (a) Probability between 0 and 1.23
1. Find the cumulative probability for \( z = 1.23 \).
2. Since we're interested in the probability between 0 and 1.23, we can calculate:
   \[
   P(0 < Z < 1.23) = P(Z < 1.23) - P(Z < 0)
   \]
3. From the standard normal table:
   - \( P(Z < 1.23) \approx 0.8907 \)
   - \( P(Z < 0) = 0.5 \)
4. Therefore:
   \[
   P(0 < Z < 1.23) = 0.8907 - 0.5 = 0.3907
   \]

### (b) Probability between −2.65 and 0
1. Find the cumulative probabilities for \( z = -2.65 \) and \( z = 0 \).
2. Calculate:
   \[
   P(-2.65 < Z < 0) = P(Z < 0) - P(Z < -2.65)
   \]
3. From the standard normal table:
   - \( P(Z < -2.65) \approx 0.0040 \)
   - \( P(Z < 0) = 0.5 \)
4. Therefore:
   \[
   P(-2.65 < Z < 0) = 0.5 - 0.0040 = 0.4960
   \]

### (c) Probability between −0.52 and 1.51
1. Find the cumulative probabilities for \( z = -0.52 \) and \( z = 1.51 \).
2. Calculate:
   \[
   P(-0.52 < Z < 1.51) = P(Z < 1.51) - P(Z < -0.52)
   \]
3. From the standard normal table:
   - \( P(Z < 1.51) \approx 0.9345 \)
   - \( P(Z < -0.52) \approx 0.3015 \)
4. Therefore:
   \[
   P(-0.52 < Z < 1.51) = 0.9345 - 0.3015 = 0.6330
   \]

### (d) Probability less than −2.62
1. Find the cumulative probability for \( z = -2.62 \).
2. Since we are looking for \( P(Z < -2.62) \), we get:
   \[
   P(Z < -2.62) \approx 0.0044
   \]

### (e) Probability greater than 2.89
1. Find the cumulative probability for \( z = 2.89 \).
2. Since we are looking for \( P(Z > 2.89) \), we calculate:
   \[
   P(Z > 2.89) = 1 - P(Z < 2.89)
   \]
3. From the standard normal table:
   - \( P(Z < 2.89) \approx 0.9981 \)
4. Therefore:
   \[
   P(Z > 2.89) = 1 - 0.9981 = 0.0019
   \]

### Summary of Answers:
- (a) \( P(0 < Z < 1.23) \approx 0.3907 \)
- (b) \( P(-2.65 < Z < 0) \approx 0.4960 \)
- (c) \( P(-0.52 < Z < 1.51) \approx 0.6330 \)
- (d) \( P(Z < -2.62) \approx 0.0044 \)
- (e) \( P(Z > 2.89) \approx 0.0019 \)

Would you like further details on any part?

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**Further Questions:**
1. What is the probability for a range not centered around zero?
2. How do we interpret these probabilities in practical terms?
3. How does changing the mean affect these probabilities?
4. Can we calculate these values without a table?
5. What are applications of normal distribution in real-world scenarios?

**Tip:** Remember that probabilities for ranges in a normal distribution can often be found by subtracting cumulative values.

Assume that the readings on the thermometers are normally distributed with a mean of 0° and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. Find the probability of each reading in degrees.

Explore how to calculate probabilities for thermometer readings normally distributed with a mean of 0° and a standard deviation of 1.00°C. This problem includes finding probabilities for specific ranges and individual values, ideal for students in Grades 10-12.

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