Mathematics, 25.03.2020 05:42 biaxialpower789

A computer battery manufacturer claims their batteries will hold a charge for an average of 15 hours, with standard deviation 3 hours. A random sample of 36 batteries is drawn. Each of the batteries is tested to see how long it holds a charge. Let X = sample mean of the 36 measured times. (a) If the claim is true, what is P(X < 14)? (b) How many batteries must be tested so that the P(X < 14) = 0.01?

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Answer from: jcazares3558

a) P(X < 14) = 0.0228.

b) 49 batteries must be tested.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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:

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

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 15, \sigma = 3, n = 36, s = \frac{3}{\sqrt{36}} = 0.5$

(a) If the claim is true, what is P(X < 14)?

This is the pvalue of Z when X = 14. So

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

By the Central Limit Theorem

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

$Z = \frac{14 - 15}{0.5}$

Z = -2

Z = -2 has a pvalue of 0.0228

P(X < 14) = 0.0228.

(b) How many batteries must be tested so that the P(X < 14) = 0.01?

Now we want to find the size of the sample for which X = 14 is in the first percentile, that is, Z when X = 14 has a pvalue of 0.01. So it is Z = -2.327.

First we have to find the standard deviation of the sample in this case.

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

$-2.327 = \frac{14 - 15}{s}$

-2.327s = -1

2.327s = 1

$s = \frac{1}{2.327}$

s = 0.43

We know that

$s = \frac{\sigma}{\sqrt{n}}$

$0.43 = \frac{3}{\sqrt{n}}$

$0.43\sqrt{n} = 3$

$\sqrt{n} = \frac{3}{0.43}$

$(\sqrt{n})^{2} = (\frac{3}{0.43})^{2}$

n = 48.7

Rounding up

49 batteries must be tested.

Answer from: Quest

d? maybe

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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