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### Given Data:
- Mean battery lifetime \( \mu = 240 \) minutes
- Standard deviation \( \sigma = 60 \) minutes
- The battery lifetime follows a normal distribution.

---

### Part (a):
**Question: What is the probability that a battery lifetime is more than 4 hours?**

First, convert 4 hours into minutes:
\[
4 \text{ hours} = 4 \times 60 = 240 \text{ minutes}.
\]
We need to calculate \( P(X > 240) \), where \( X \) is the battery lifetime.

Since \( X \) follows a normal distribution with \( \mu = 240 \) and \( \sigma = 60 \), the probability of \( X > 240 \) corresponds to:
\[
P(X > 240) = P(Z > 0),
\]
where \( Z \) is the standard normal variable. The probability of \( Z > 0 \) is 0.5.

So, the probability that the battery lifetime is more than 4 hours is:
\[
P(X > 240) = 0.5.
\]

---

### Part (b):
**Question: The quartiles (the 25% and 75% values) of battery lifetime are:**

The 25th and 75th percentiles correspond to the following \( z \)-scores in the standard normal distribution:
- For the 25th percentile: \( z_{25\%} = -0.674 \).
- For the 75th percentile: \( z_{75\%} = 0.674 \).

Using the formula for converting a \( z \)-score to the corresponding \( x \)-value in a normal distribution:
\[
x = \mu + z \cdot \sigma,
\]
we can calculate:
- For the 25th percentile:
  \[
  x_{25\%} = 240 + (-0.674) \times 60 \approx 240 - 40.44 = 199.56 \text{ minutes}.
  \]
- For the 75th percentile:
  \[
  x_{75\%} = 240 + 0.674 \times 60 \approx 240 + 40.44 = 280.44 \text{ minutes}.
  \]

So, the 25% value is approximately 200 minutes, and the 75% value is approximately 280 minutes.

---

### Part (c):
**Question: What value of battery lifetime is exceeded with a probability of 0.95?**

To find the value of \( x \) such that \( P(X > x) = 0.95 \), we need to look up the corresponding \( z \)-score for a cumulative probability of \( 1 - 0.95 = 0.05 \).

The \( z \)-score for \( P(Z > z) = 0.05 \) is approximately \( z = -1.645 \).

Now, use the formula:
\[
x = \mu + z \cdot \sigma,
\]
\[
x = 240 + (-1.645) \times 60 \approx 240 - 98.7 = 141.3 \text{ minutes}.
\]

Therefore, the value of battery lifetime exceeded with 95% probability is approximately 141 minutes.

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Here are 5 questions based on this information:
1. How would the probabilities change if the standard deviation increased?
2. What is the interpretation of the quartiles in this context?
3. How would you calculate the probability that the battery lasts between 3 and 5 hours?
4. What does the 95th percentile represent in this scenario?
5. How would the probabilities shift if the mean battery lifetime increased by 30 minutes?

**Tip:** Understanding how to calculate the probability from a normal distribution can be useful in many fields where data follows a normal pattern.

The time until a battery in a laptop computer is fully depleted under common conditions is normally distributed with a mean lifetime of 240 minutes and a standard deviation of 60. (a) What is the probability that a battery lifetime is more than 4 hours? (b) The quartiles (the 25% and 75% values) of battery lifetime are: (c) What value of battery lifetime is exceeded with probability 0.95?

Solve the probability problem of a battery's lifetime using normal distribution with a mean of 240 minutes and a standard deviation of 60 minutes. Learn how to calculate probabilities, quartiles, and exceedance values using standard normal distribution techniques.

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