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Sagot :
To find the probability that a given light bulb lasts between 525 and 750 hours, given that the life of light bulbs follows a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 750 hours and a standard deviation ([tex]\(\sigma\)[/tex]) of 75 hours, we can use the empirical rule, often known as the [tex]\(68 \%-95 \%-99.7 \%\)[/tex] rule.
### Step-by-Step Solution
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]): 750 hours
- Standard deviation ([tex]\(\sigma\)[/tex]): 75 hours
- Lower bound: 525 hours
- Upper bound: 750 hours
2. Convert the given bounds to Z-scores:
The Z-score is calculated using the formula:
[tex]\[ Z = \frac{(X - \mu)}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value of interest.
- For the lower bound (525 hours):
[tex]\[ Z_{\text{lower}} = \frac{(525 - 750)}{75} = \frac{-225}{75} = -3.0 \][/tex]
- For the upper bound (750 hours):
[tex]\[ Z_{\text{upper}} = \frac{(750 - 750)}{75} = \frac{0}{75} = 0.0 \][/tex]
3. Apply the [tex]\(68\%-95\%-99.7\%\)[/tex] rule:
The [tex]\(68\%-95\%-99.7\%\)[/tex] rule states that for a normal distribution:
- Approximately [tex]\(68\%\)[/tex] of the data lies within one standard deviation ([tex]\(\mu \pm \sigma\)[/tex]) from the mean.
- Approximately [tex]\(95\%\)[/tex] of the data lies within two standard deviations ([tex]\(\mu \pm 2\sigma\)[/tex]) from the mean.
- Approximately [tex]\(99.7\%\)[/tex] of the data lies within three standard deviations ([tex]\(\mu \pm 3\sigma\)[/tex]) from the mean.
In this case:
- The interval [tex]\(525 \leq b \leq 750\)[/tex] corresponds to Z-scores of [tex]\(-3\)[/tex] to [tex]\(0\)[/tex], which spans from the mean (750) to [tex]\(3\sigma\)[/tex] below the mean (525).
4. Determine the probability:
Given that [tex]\(68\%\)[/tex] of the data lies within one standard deviation ([tex]\(\mu \pm \sigma\)[/tex]), we can deduce that half of this percentage (from the mean to [tex]\(\mu - \sigma\)[/tex]) is [tex]\(34\%\)[/tex]. Therefore, the probability that a light bulb lasts between 525 and 750 hours is [tex]\(34\%\)[/tex] or [tex]\(0.34\)[/tex].
### Summary
Thus, the probability that a given light bulb lasts between 525 and 750 hours is approximately [tex]\(0.34\)[/tex] or [tex]\(34\%\)[/tex]. This conclusion follows from the empirical [tex]\(68\%-95\%-99.7\%\)[/tex] when the provided Z-scores were calculated and evaluated.
### Step-by-Step Solution
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]): 750 hours
- Standard deviation ([tex]\(\sigma\)[/tex]): 75 hours
- Lower bound: 525 hours
- Upper bound: 750 hours
2. Convert the given bounds to Z-scores:
The Z-score is calculated using the formula:
[tex]\[ Z = \frac{(X - \mu)}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value of interest.
- For the lower bound (525 hours):
[tex]\[ Z_{\text{lower}} = \frac{(525 - 750)}{75} = \frac{-225}{75} = -3.0 \][/tex]
- For the upper bound (750 hours):
[tex]\[ Z_{\text{upper}} = \frac{(750 - 750)}{75} = \frac{0}{75} = 0.0 \][/tex]
3. Apply the [tex]\(68\%-95\%-99.7\%\)[/tex] rule:
The [tex]\(68\%-95\%-99.7\%\)[/tex] rule states that for a normal distribution:
- Approximately [tex]\(68\%\)[/tex] of the data lies within one standard deviation ([tex]\(\mu \pm \sigma\)[/tex]) from the mean.
- Approximately [tex]\(95\%\)[/tex] of the data lies within two standard deviations ([tex]\(\mu \pm 2\sigma\)[/tex]) from the mean.
- Approximately [tex]\(99.7\%\)[/tex] of the data lies within three standard deviations ([tex]\(\mu \pm 3\sigma\)[/tex]) from the mean.
In this case:
- The interval [tex]\(525 \leq b \leq 750\)[/tex] corresponds to Z-scores of [tex]\(-3\)[/tex] to [tex]\(0\)[/tex], which spans from the mean (750) to [tex]\(3\sigma\)[/tex] below the mean (525).
4. Determine the probability:
Given that [tex]\(68\%\)[/tex] of the data lies within one standard deviation ([tex]\(\mu \pm \sigma\)[/tex]), we can deduce that half of this percentage (from the mean to [tex]\(\mu - \sigma\)[/tex]) is [tex]\(34\%\)[/tex]. Therefore, the probability that a light bulb lasts between 525 and 750 hours is [tex]\(34\%\)[/tex] or [tex]\(0.34\)[/tex].
### Summary
Thus, the probability that a given light bulb lasts between 525 and 750 hours is approximately [tex]\(0.34\)[/tex] or [tex]\(34\%\)[/tex]. This conclusion follows from the empirical [tex]\(68\%-95\%-99.7\%\)[/tex] when the provided Z-scores were calculated and evaluated.
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