Question 1182194: Q.1 Global financial institution transfers a large data file every evening form offices around the world to its London headquarters. Once the file is received, it must be cleaned and partitioned before being stored in the company’s data warehouse. Each file is the same size and the time required to transfer, clean, and partition a file is normally distributed, with a mean of 1.5 hours and a standard deviation of 15 minutes.
a. if one file is selected at random, what is the probability that it will take longer than 1 hour and 55 minutes to transfer, clean, and partition the file?
b. If a manager must be present until 85% of the files are transferred, cleaned, and partitioned, how long will the manager need to be there?
c. What percentage of the data files will take between 63 minutes and 110 minutes to be transferred, cleaned, and partitioned?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**a. Probability of taking longer than 1 hour and 55 minutes:**
1. **Convert to a common unit:** Let's work with minutes. 1 hour and 55 minutes is equal to 115 minutes. The mean time is 1.5 hours * 60 minutes/hour = 90 minutes.
2. **Calculate the z-score:** The z-score tells us how many standard deviations a value is from the mean.
z = (x - μ) / σ
z = (115 - 90) / 15
z = 1.67
3. **Find the probability:** Use a standard normal distribution table (a z-table) or a calculator to find the probability associated with a z-score of 1.67. We want the probability of the time being *greater* than 115 minutes, so we look for the area to the *right* of z = 1.67.
P(x > 115) = P(z > 1.67) = 1 - P(z < 1.67) ≈ 1 - 0.9525 ≈ 0.0475
**b. Time for the manager to be present until 85% of files are processed:**
1. **Find the z-score:** We want to find the time *t* such that 85% of the files are processed in less than *t* minutes. This corresponds to a cumulative probability of 0.85. Look up 0.85 in the *body* of the z-table (or use a calculator) to find the corresponding z-score. z ≈ 1.04
2. **Convert the z-score to time:**
x = μ + zσ
x = 90 + (1.04 * 15)
x ≈ 105.6 minutes
**c. Percentage of files taking between 63 and 110 minutes:**
1. **Calculate the z-scores:**
z₁ = (63 - 90) / 15 = -1.8
z₂ = (110 - 90) / 15 = 1.33
2. **Find the probabilities:** Use a z-table or calculator.
P(x < 63) = P(z < -1.8) ≈ 0.0359
P(x < 110) = P(z < 1.33) ≈ 0.9082
3. **Calculate the probability between the two values:**
P(63 < x < 110) = P(z < 1.33) - P(z < -1.8) ≈ 0.9082 - 0.0359 ≈ 0.8723
4. **Convert to percentage:** 0.8723 * 100% ≈ 87.23%
**Answers:**
* a. The probability that a randomly selected file will take longer than 1 hour and 55 minutes is approximately 0.0475 or 4.75%.
* b. The manager needs to be present for approximately 105.6 minutes.
* c. Approximately 87.23% of the data files will take between 63 minutes and 110 minutes to be transferred, cleaned, and partitioned.
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