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tresset_1 [31]
3 years ago
10

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places.
Mathematics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is \\ P(z>0.48) = P(x>533.2) = 0.3156

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> with mean \\ \mu and standard deviation \\ \frac{\sigma}{\sqrt{n}}. Mathematically

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

For values of the sample \\ n \ge 30, no matter the distribution the data come from.

And the variable <em>z</em> follows a <em>standard normal distribution</em>, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} [1]

That is

\\ z \sim N(0, 1)

We have a variance of 3364. That is, a <em>standard deviation</em> of

\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58

The population mean is

\\ \mu = 530

The sample size is \\ n = 75

The sample mean is \\ \overline{x} = 533.2

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}

\\ z = \frac{3.2}{\frac{58}{8.66025}}

\\ z = \frac{3.2}{6.69726}

\\ z = 0.47780

With this value of z we can consult a <em>cumulative standard normal table</em> (or use some statistic program) to find the cumulative probability for <em>z</em> (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

\\ P(z

However, in the question we are asked for \\ P(z>0.48) = P(x>533.2). As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

\\ P(z>0.48) = 1 - P(z

Thus

\\ P(z>0.48) = 1 - 0.68439

\\ P(z>0.48) = 0.31561

Rounding to four decimal places, we have

\\ P(z>0.48) = 0.3156

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) \\ P(z>0.48) = P(x>533.2) = 0.3156.

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