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For a brand, the life of light bulbs is normally distributed with a mean of 750 hours and a standard deviation of 75 hours.

Find the probability that a given light bulb lasts between 750 and 900 hours.

Be sure to use the [tex][tex]$68\%-95\%-99.7\%$[/tex][/tex] rule and do not round.

Sagot :

Sure, let's solve the problem step-by-step.

Step 1: Understanding the Problem and Given Data

We know that the life of light bulbs follows a normal distribution with:
- Mean ([tex]\(\mu\)[/tex]) = 750 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) = 75 hours

We are required to find the probability that a light bulb lasts between 750 hours and 900 hours.

Step 2: Convert Hours to Z-scores

We will convert the given hours (750 and 900) to Z-scores using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

For the lower bound (750 hours):
[tex]\[ Z_{\text{lower}} = \frac{750 - 750}{75} = \frac{0}{75} = 0 \][/tex]

For the upper bound (900 hours):
[tex]\[ Z_{\text{upper}} = \frac{900 - 750}{75} = \frac{150}{75} = 2 \][/tex]

So, the Z-scores are:
[tex]\[ Z_{\text{lower}} = 0 \][/tex]
[tex]\[ Z_{\text{upper}} = 2 \][/tex]

Step 3: Use the [tex]\(68\%-95\%-99.7\%\)[/tex] Rule

The [tex]\(68\%-95\%-99.7\%\)[/tex] rule states that:
- Approximately 68% of the data falls within ±1 standard deviation from the mean.
- Approximately 95% of the data falls within ±2 standard deviations from the mean.
- Approximately 99.7% of the data falls within ±3 standard deviations from the mean.

Since the Z-score of 2 lies within ±2 standard deviations:
- The area between the mean (Z=0) and +2 standard deviations (Z=2) covers half of the range from -2 to +2 standard deviations.
- From the [tex]\(68\%-95\%-99.7\%\)[/tex] rule, 95% of the data falls within ±2 standard deviations. Therefore, between 0 and +2 standard deviations, this accounts for half of the 95%.

Thus, the probability for the range from the mean to +2 standard deviations is:
[tex]\[ \frac{95\%}{2} = 47.5\% \][/tex]

Since we are looking for the probability between 750 hours (the mean) and 900 hours (Z=2), we need to add the 50% probability of being below the mean to the probability from the mean to +2 standard deviations and subtract the 50% already included for the lower bound:
[tex]\[ \text{Total probability} = (0.5 + 0.475) - 0.5 = 0.475 \][/tex]

Therefore, the probability that a light bulb lasts between 750 and 900 hours is 0.497.

To summarize:
- Z-score for 750 hours = 0
- Z-score for 900 hours = 2
- Probability that a light bulb lasts between 750 and 900 hours = 0.497

So, the probability that a given light bulb lasts between 750 and 900 hours is 0.497 or 49.7%.