Question

The lifetime of lightbulbs that are advertised to last for 4000 hours are normally distributed with a mean of 4250 hours and a standard deviation of 300 hours. What is the probability that a bulb lasts longer than advertised

Answer 1

Answer:

79.67% probability that a bulb lasts longer than advertised

Step-by-step explanation:

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

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 4250, \sigma = 300$

What is the probability that a bulb lasts longer than advertised?

This is 1 subtracted by the pvalue of Z when X = 4000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4000 - 4250}{300}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033

1 - 0.2033 = 0.7967

79.67% probability that a bulb lasts longer than advertised