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To determine the probability that the lifespan of a randomly chosen hard drive from this manufacturer lies between 2 years 3 months and 3 years 3 months, we need to convert these time periods into their standard normal z-scores, and then use the standard normal distribution table to find corresponding probabilities.
### Step-by-Step Solution
#### 1. Convert Lifespan Units to Years
Firstly, we convert the given lifespan durations to years for consistent calculations:
- Mean lifespan: 3 years and 6 months = \( 3 + \frac{6}{12} = 3.5 \) years
- Standard deviation: 9 months = \( \frac{9}{12} = 0.75 \) years
- Lower bound: 2 years and 3 months = \( 2 + \frac{3}{12} = 2.25 \) years
- Upper bound: 3 years and 3 months = \( 3 + \frac{3}{12} = 3.25 \) years
#### 2. Calculate Z-Scores
The z-scores are calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
Where:
- \( X \) is the value for which we are finding the z-score,
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.
For the lower bound (2.25 years):
[tex]\[ z_{\text{lower}} = \frac{2.25 - 3.5}{0.75} = \frac{-1.25}{0.75} \approx -1.67 \][/tex]
For the upper bound (3.25 years):
[tex]\[ z_{\text{upper}} = \frac{3.25 - 3.5}{0.75} = \frac{-0.25}{0.75} \approx -0.33 \][/tex]
#### 3. Use the Standard Normal Table
Using the provided portion of the standard normal table, we find the probabilities associated with each z-score.
- For \( z_{\text{lower}} \approx -1.67 \):
Since the standard normal distribution table typically gives the area to the left of the z-score, and given that it is symmetric, the probability corresponding to \( z = -1.67 \) is \( 1 - 0.9525 = 0.0475 \).
- For \( z_{\text{upper}} \approx -0.33 \):
Using the standard normal table, the probability corresponding to \( z = -0.33 \) is found by subtracting the area to the left of \( z = 0.33 \) from 1 to account for negative z-values. Using symmetry and looking up \( z = 0.33 \), we see \( 1 - 0.6293 = 0.3707 \).
#### 4. Calculate the Desired Probability
The probability that the lifespan lies between the lower and upper bounds is found by subtracting the cumulative probability at the lower bound from the cumulative probability at the upper bound:
[tex]\[ P(2.25 < X < 3.25) = P(z_{\text{upper}}) - P(z_{\text{lower}}) = 0.3707 - 0.0475 = 0.3232 \][/tex]
### Conclusion
Thus, the probability that a randomly selected hard drive from the manufacturer lasts between 2 years 3 months and 3 years 3 months is approximately [tex]\(32.32\%\)[/tex].
### Step-by-Step Solution
#### 1. Convert Lifespan Units to Years
Firstly, we convert the given lifespan durations to years for consistent calculations:
- Mean lifespan: 3 years and 6 months = \( 3 + \frac{6}{12} = 3.5 \) years
- Standard deviation: 9 months = \( \frac{9}{12} = 0.75 \) years
- Lower bound: 2 years and 3 months = \( 2 + \frac{3}{12} = 2.25 \) years
- Upper bound: 3 years and 3 months = \( 3 + \frac{3}{12} = 3.25 \) years
#### 2. Calculate Z-Scores
The z-scores are calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
Where:
- \( X \) is the value for which we are finding the z-score,
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.
For the lower bound (2.25 years):
[tex]\[ z_{\text{lower}} = \frac{2.25 - 3.5}{0.75} = \frac{-1.25}{0.75} \approx -1.67 \][/tex]
For the upper bound (3.25 years):
[tex]\[ z_{\text{upper}} = \frac{3.25 - 3.5}{0.75} = \frac{-0.25}{0.75} \approx -0.33 \][/tex]
#### 3. Use the Standard Normal Table
Using the provided portion of the standard normal table, we find the probabilities associated with each z-score.
- For \( z_{\text{lower}} \approx -1.67 \):
Since the standard normal distribution table typically gives the area to the left of the z-score, and given that it is symmetric, the probability corresponding to \( z = -1.67 \) is \( 1 - 0.9525 = 0.0475 \).
- For \( z_{\text{upper}} \approx -0.33 \):
Using the standard normal table, the probability corresponding to \( z = -0.33 \) is found by subtracting the area to the left of \( z = 0.33 \) from 1 to account for negative z-values. Using symmetry and looking up \( z = 0.33 \), we see \( 1 - 0.6293 = 0.3707 \).
#### 4. Calculate the Desired Probability
The probability that the lifespan lies between the lower and upper bounds is found by subtracting the cumulative probability at the lower bound from the cumulative probability at the upper bound:
[tex]\[ P(2.25 < X < 3.25) = P(z_{\text{upper}}) - P(z_{\text{lower}}) = 0.3707 - 0.0475 = 0.3232 \][/tex]
### Conclusion
Thus, the probability that a randomly selected hard drive from the manufacturer lasts between 2 years 3 months and 3 years 3 months is approximately [tex]\(32.32\%\)[/tex].
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