The lifetime of battery of a device (in one charge) is normally distributed with mean µ = 18 hours and standard deviation σ = 2 hours. (a) What is the probability that a battery will last more than 20 hours? (b) Find the 10th percentile of the lifetimes. (c) A particular battery lasts 16 hours. What percentile is its lifetime on? (d) What is the probability that the lifetime of a battery is between 17.5 and 18.5? (e) Ten batteries are chosen at random, what is the probability that the mean lifetime is between 17.5 and 18.5?

Answer:

a) 0.1587 = 15.87% probability that a battery will last more than 20 hours.

b) 15.44 hours.

c) Approximately the 16th percentile.

d) 0.1974 = 19.74% probability that the lifetime of a battery is between 17.5 and 18.5.

e) 0.5704 = 57.04% probability that the mean lifetime is between 17.5 and 18.5

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this question, we have that:

[tex]\mu = 18, \sigma = 2[/tex]

(a) What is the probability that a battery will last more than 20 hours?

This is 1 subtracted by the pvalue of Z when X = 20. So

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

[tex]Z = \frac{20 - 18}{2}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413

1 - 0.8413 = 0.1587

0.1587 = 15.87% probability that a battery will last more than 20 hours.

(b) Find the 10th percentile of the lifetimes.

This is X when Z has a pvalue of 0.1. So X when Z = -1.28

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

[tex]-1.28 = \frac{X - 18}{2}[/tex]

[tex]X - 18 = -1.28*2[/tex]

[tex]X = 15.44[/tex]

So 15.44 hours.

(c) A particular battery lasts 16 hours. What percentile is its lifetime on?

We have to find the pvalue of Z when X = 16. So

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

[tex]Z = \frac{16 - 18}{2}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

Approximately the 16th percentile.

(d) What is the probability that the lifetime of a battery is between 17.5 and 18.5?

This is the pvalue of Z when X = 18.5 subtracted by the pvalue of Z when X = 17.5. So

X = 18.5

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

[tex]Z = \frac{18.5 - 18}{2}[/tex]

[tex]Z = 0.25[/tex]

[tex]Z = 0.25[/tex] has a pvalue of 0.5987

X = 17.5

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

[tex]Z = \frac{17.5 - 18}{2}[/tex]

[tex]Z = -0.25[/tex]

[tex]Z = -0.25[/tex] has a pvalue of 0.4013

0.5987 - 0.4013 = 0.1974

0.1974 = 19.74% probability that the lifetime of a battery is between 17.5 and 18.5.

(e) Ten batteries are chosen at random, what is the probability that the mean lifetime is between 17.5 and 18.5?

Sample of 10 means that, by the Central Limit Theorem, [tex]n = 10, s = \frac{2}{\sqrt{10}} = 0.6325[/tex]

X = 18.5

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

[tex]Z = \frac{18.5 - 18}{0.6325}[/tex]

[tex]Z = 0.79[/tex]

[tex]Z = 0.79[/tex] has a pvalue of 0.7852

X = 17.5

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

[tex]Z = \frac{17.5 - 18}{0.6325}[/tex]

[tex]Z = -0.79[/tex]

[tex]Z = -0.79[/tex] has a pvalue of 0.2148

0.7852 - 0.2148 = 0.5704

0.5704 = 57.04% probability that the mean lifetime is between 17.5 and 18.5

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