Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 5929 with a mean life of 941 minutes. If the claim is true, in a sample of 109 batteries, what is the probability that the mean battery life would be greater than 948.8 minutes? Round your answer to four decimal places.

Answer 1

Answer:

"The probability that the mean battery life would be greater than 948.8 minutes" is 0.1446.

Step-by-step explanation:

In this case, the quality control expert takes a sample of batteries. From these batteries, we want to find "the probability that the mean battery life would be greater than 948.8 minutes".

Different concepts needed to take into account to solve this question

Sampling Distribution of the Means

For doing this, we need to use the sampling distribution of the means, which results from taking the mean for each possible sample coming from a random variable $\\ x$. Roughly speaking, each sample will have a different mean, $\\ \overline{x}$, and the probability distribution for any of these means is called the sampling distribution of the means.

The sampling distribution of the means has a mean that equals the population's mean for the random variable $\\ x$, i.e., $\\ \mu$, and its standard deviation is $\\ \frac{\sigma}{\sqrt{n}}$. We can express this mathematically as:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

Standardized Values for $\\ \overline{x}$

We can standardized the values for $\\ \overline{x}$ using z-scores:

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

This random variable $\\ Z$ follows a standard normal distribution, that is, $\\ Z \sim N(0,1)$, and it is easier to find probabilities since the values for them are tabulated in the standard normal table (available in any Statistics book or on the Internet.)

What type of distribution follows the sampling distribution of the means?

A general rule of thumb is that this distribution (the sampling distribution of the means) follows a normal distribution if the sample size, $\\ n$, is bigger than or equal to 30 observations, or $\\ n \geq 30$. In this case, $\\ n = 109$ batteries. This is a result from the Central Limit Theorem, fundamental in Statistical Inference.

Standard Deviation

We have to remember that the standard deviation is the square root of the variance $\\ \sigma^2$, or $\\ \sqrt{\sigma^2}$.

• $\\ \sigma^{2} =5929$
• $\\ \sigma = \sqrt{5929} = 77$

Therefore, the standard deviation in this case is $\\ \sigma = 77$ minutes.

In sum, we have the following information to answer this question:

• $\\ \sigma = 77$ minutes.
• $\\ \mu = 941$ minutes.
• $\\ n = 109$ batteries (the sample size is large enough to assume that the sampling distribution of the means follows a normal distribution).
• $\\ \overline{x} = 948.8$ minutes.

What is the probability that the mean battery life would be greater than 948.8 minutes?

Well, having all the previous information, we can use [2] to solve this question (without using units):

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

$\\ z = \frac{948.8 - 941}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{7.37526}$

$\\ z = 1.05758 \approx 1.06$

This result is the standardized value or z-score for $\\ \overline{x}$, considering $\\ \mu = 941$ and $\\ \sigma = 77$.

We round z to two decimals digits since standard normal table only uses it as an entry to find probabilities.

With $\\ z = 1.06$, we can consult the cumulative standard normal table. First, we need to find with $\\ z = 1.0$ in the first column in the table. Then, in its first raw, we need to find +0.06. The intersection for these two values determines the cumulative probability for $\\ P(z$.

It is important to recall that $\\ P(z$ because $\\ z = 1.06$ is the standardized value for $\\ \overline{x} = 948.8$ minutes.

Then,  $\\ P(z$

However, the question is about $\\ P(\overline{x} > 948.8) = P(z>1.06)$

And

$\\ P(\overline{x} > 948.8) + P(\overline{x} < 948.8) = 1$

Or

$\\ P(z>1.06) + P(z$

Then

$\\ P(z>1.06) = 1 - P(z$

$\\ P(z>1.06) = 1 - 0.8554$

$\\ P(z>1.06) = 0.1446$

Therefore, "the probability that the mean battery life would be greater than 948.8 minutes" is 0.1446.