A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.1 years, and standard deviation of 1.3 years. If you randomly purchase one item, what is the probability it will last longer than 7 years

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Answer:

24.51% probability it will last longer than 7 years

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 problem, we have that:

[tex]\mu = 6.1, \sigma = 1.3[/tex]

If you randomly purchase one item, what is the probability it will last longer than 7 years

This is 1 subtracted by the pvalue of Z when X = 7. So

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

[tex]Z = \frac{7 - 6.1}{1.3}[/tex]

[tex]Z = 0.69[/tex]

[tex]Z = 0.69[/tex] has a pvalue of 0.7549

1 - 0.7549 = 0.2451

24.51% probability it will last longer than 7 years