# The lifetimes of projector bulbs of a particular type are normally distributed with a mean of 470 hours and a standard deviation of 15 hours. What percentage of the bulbs has lifetimes that lie within 2 standard deviations of the mean on either side?

## To find the percentage of projector bulbs that have lifetimes within 2 standard deviations of the mean, we can use the properties of the normal distribution.

Step 1: Calculate the values of the boundaries.
The mean of the distribution is given as 470 hours, and the standard deviation is given as 15 hours. The boundaries are defined as 2 standard deviations away from the mean on either side.

So, the lower boundary is:
Lower boundary = mean - 2 * standard deviation
= 470 - 2 * 15
= 470 - 30
= 440 hours

And the upper boundary is:
Upper boundary = mean + 2 * standard deviation
= 470 + 2 * 15
= 470 + 30
= 500 hours

Step 2: Calculate the area under the normal distribution curve between these boundaries.
The area under the normal distribution curve represents the probability.

To find this probability using a standard normal distribution table, we need to calculate the z-scores for the lower and upper boundaries.

The z-score is calculated using the formula:
z = (x - mean) / standard deviation

For the lower boundary:
z_lower = (440 - 470) / 15
= -30 / 15
= -2

For the upper boundary:
z_upper = (500 - 470) / 15
= 30 / 15
= 2

Using the z-scores, we can find the corresponding probabilities from the standard normal distribution table.

Step 3: Look up the z-scores in the standard normal distribution table to find the corresponding probabilities.
The cumulative probability for a given z-score represents the percentage of values that are less than or equal to that z-score.

From the table, the cumulative probability for z = -2 is approximately 0.0228, and the cumulative probability for z = 2 is approximately 0.9772.

Step 4: Calculate the percentage of bulbs that lie within 2 standard deviations of the mean.
Since we want to find the percentage of bulbs between the lower boundary and the upper boundary, we subtract the cumulative probability corresponding to the lower boundary from the cumulative probability corresponding to the upper boundary.

Percentage = (Cumulative probability for upper boundary) - (Cumulative probability for lower boundary)
= 0.9772 - 0.0228
= 0.9544

Therefore, approximately 95.44% of the bulbs have lifetimes that lie within 2 standard deviations of the mean on either side.