A new extended life light bulb has an average service life of 750 hours, with a standard deviation of 50 hours. If the service life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 600 hours and 900 hours?
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z = (X-u)/sigma. z1 = (600 - 750)/50 = -3.0. z2 = (900 - 750)/50 = 3.0. from the z tables, z = 3 is 0.4987. Therefore the area between z1 and z2 is 0.4987*2 =
0.9974
To determine the percentage of the distribution that falls between 600 hours and 900 hours, we need to calculate the z-scores for these two values and then use the standard normal distribution table.
First, we calculate the z-score for 600 hours using the formula: z = (x - μ) / σ where x is the given value, μ is the mean, and σ is the standard deviation.
For 600 hours: z = (600 - 750) / 50 z = -3
Next, we calculate the z-score for 900 hours: z = (900 - 750) / 50 z = 3
Now, we look up the corresponding areas under the standard normal distribution curve for z-scores of -3 and 3. These values represent the percentage of the distribution that falls below those z-scores.
From the standard normal distribution table, we find that the area to the left of a z-score of -3 is approximately 0.0013 (or 0.13%). This means that about 0.13% of the distribution falls below 600 hours.
Similarly, the area to the left of a z-score of 3 is approximately 0.9987 (or 99.87%). This means that about 99.87% of the distribution falls below 900 hours.
To find the percentage of the distribution between 600 hours and 900 hours, we subtract the percentage below 600 hours from the percentage below 900 hours: Percentage = 99.87% - 0.13% Percentage = 99.74%
Therefore, approximately 99.74% of the distribution will be between 600 hours and 900 hours.
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