The life time of a certain brand of bulbs produced by a company is normally distributed, with mean 210 hours and standard deviation 56 hours.what is the probability that a bulb picked at random from this company product will have a life time of at least 300 hours?

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Answer:

5.37% probability that a bulb picked at random from this company product will have a life time of at least 300 hours

Step-by-step explanation:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 question, we have that:

[tex]\mu = 210, \sigma = 56[/tex]

What is the probability that a bulb picked at random from this company product will have a life time of at least 300 hours?

This is 1 subtracted by the pvalue of Z when X = 300. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{300 - 210}{56}[/tex]

[tex]Z = 1.61[/tex]

[tex]Z = 1.61[/tex] has a pvalue of 0.9463

1 - 0.9463 = 0.0537

5.37% probability that a bulb picked at random from this company product will have a life time of at least 300 hours