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atroni [7]
3 years ago
12

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There

are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?
Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

The lifetime value needed is 11.8225 hours.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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 This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that \mu = 11, \sigma = 1.

What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use Z = 1.645

Since we need the mean sample, we need to find the standard deviation of the sample, that is:

s = \frac{\sigma}{\sqrt{4}} = 0.5

So:

Z = \frac{X - \mu}{s}

1.645 = \frac{X - 11}{0.5}

X - 11 = 0.5*1.645

X = 11.8225

The lifetime value needed is 11.8225 hours.

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