Question

A particular battery claims to have a mean life of 400 hours with a standard deviation of 30 hours. Approximately what percent of the batteries will last more than 420 hours?

Answer 1

Using the normal distribution, it is found that 25.14% of the batteries will last more than 420 hours.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, we have that the mean and the standard deviation are given, respectively, by:

$\mu = 400, \sigma = 30$.

The proportion of the batteries will last more than 420 hours is one subtracted by the p-value of Z when X = 420, hence:

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

$Z = \frac{420 - 400}{30}$

Z = 0.67

Z = 0.67 has a p-value of 0.7486.

1 - 0.7486 = 0.2514.

0.2514 = 25.14% of the batteries will last more than 420 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

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