SOLUTION: The length of time in hours before a mobile requires charging has a normal distribution with a mean of 100 hours and a standard deviation of 15 hours. a) Find the probability tha

Question 1182183: The length of time in hours before a mobile requires charging has a normal distribution with a mean of
100 hours and a standard deviation of 15 hours.
a) Find the probability that the time before charging is greater than 127 hours.
b) Find the 10th percentile
c) You are about to go on a 6 hour trip. Given you last charged your phone 127 hours ago, what is
the probability your mobile will not need charging until you complete the trip?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**a) Probability of time before charging being greater than 127 hours:**
1. **Calculate the z-score:**
z = (x - μ) / σ
z = (127 - 100) / 15
z = 1.8
2. **Find the probability:** Use a standard normal distribution table (z-table) or a calculator to find the probability associated with a z-score of 1.8. We want the probability of the time being *greater* than 127 hours, so we look for the area to the *right* of z = 1.8.
P(x > 127) = P(z > 1.8) = 1 - P(z < 1.8) ≈ 1 - 0.9641 ≈ 0.0359
**b) Find the 10th percentile:**
1. **Find the z-score:** The 10th percentile is the value below which 10% of the data falls. Look up 0.10 in the *body* of the z-table (or use a calculator) to find the corresponding z-score. The z-score corresponding to 0.10 is approximately -1.28.
2. **Convert the z-score to hours:**
x = μ + zσ
x = 100 + (-1.28 * 15)
x ≈ 80.8 hours
**c) Probability of not needing charging during a 6-hour trip, given it was charged 127 hours ago:**
This is a conditional probability problem. We know the phone lasted 127 hours, and we want to know the probability it will last *at least* an additional 6 hours (127 + 6 = 133 hours).
1. **Calculate the conditional probability:** We want P(x > 133 | x > 127). This can be written as:
P(x > 133 and x > 127) / P(x > 127)
Since if x>133, it is automatically true that x>127, then P(x > 133 and x > 127) simplifies to P(x>133)
So we have P(x > 133) / P(x > 127)
2. **Calculate P(x > 133):**
z = (133 - 100) / 15
z = 2.2
P(x > 133) = P(z > 2.2) = 1 - P(z < 2.2) ≈ 1 - 0.9861 ≈ 0.0139
3. **Calculate the conditional probability:**
P(x > 133 | x > 127) = P(x > 133) / P(x > 127) ≈ 0.0139 / 0.0359 ≈ 0.387
**Answers:**
* a) The probability that the time before charging is greater than 127 hours is approximately 0.0359 or 3.59%.
* b) The 10th percentile is approximately 80.8 hours.
* c) The probability that the mobile will not need charging during the 6-hour trip, given it was charged 127 hours ago, is approximately 0.387 or 38.7%.

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