A certain brand of lightbulb is advertised to last for 6500 hours. The life-span of a randomly selected bulb is normally distributed with a mean of 6600 hours and a standard deviation of 200 hours. What is the probability that a bulb lasts longer than its advertised life-span?

69.15% probability that a bulb lasts longer than its advertised life-span

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What is the probability that a bulb lasts longer than its advertised life-span?

Advertised to last 6500 hours, so this probability is 1 subtracted by the pvalue of Z when X = 6500.

has a pvalue of 0.3085.

1 - 0.3085 = 0.6915

69.15% probability that a bulb lasts longer than its advertised life-span

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: