The life of light bulbs is distributed normally. The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours. Find the probability of a bulb lasting for at most 528 hours. Round your answer to four decimal places.

0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours.

This means that

Find the probability of a bulb lasting for at most 528 hours.

This is the p-value of Z when X = 528. So

has a p-value of 0.7031

0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

the radius is 2 times the diameter.

step-by-step explanation: