Mattel Corporation produces a remote-controlled car that requires three AA batteries. The mean life of these batteries in this product is 34 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of their testing program Sony tests samples of 25 batteries.

Question

a.) What is the standard error of the distribution of the sample mean?
b.) What is the probability that the sample mean life is between 34.5 and 36.0
hours

Answer 1

To solve the problems stated, we will use the following concepts from statistics related to normal distributions and sample means.

a) Standard Error of the Distribution of the Sample Mean

The standard error (SE) of the sample mean is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} \]

where:

• \(\sigma\) is the population standard deviation,
• \(n\) is the sample size.

Given:

• \(\sigma = 5.5\) hours,
• \(n = 25\),

Substituting the values in:

\[ SE = \frac{5.5}{\sqrt{25}} = \frac{5.5}{5} = 1.1 \text{ hours} \]

So, the standard error of the distribution of the sample mean is 1.1 hours.

b) Probability that the Sample Mean Life is Between 34.5 and 36.0 Hours

To find the probability that the sample mean lies between 34.5 and 36.0 hours, we will calculate the z-scores for both values using the formula:

\[ z = \frac{\bar{x} - \mu}{SE} \]

where:

• \(\bar{x}\) is the sample mean,
• \(\mu\) is the population mean (34 hours in this case),
• \(SE\) is the standard error we calculated earlier (1.1 hours).

1. Calculate the z-score for 34.5 hours:

\[ z_{34.5} = \frac{34.5 - 34}{1.1} = \frac{0.5}{1.1} \approx 0.4545 \]

2. Calculate the z-score for 36.0 hours:

\[ z_{36} = \frac{36 - 34}{1.1} = \frac{2}{1.1} \approx 1.8182 \]

Now we need to use the standard normal distribution to find the probabilities corresponding to these z-scores.

3. Look up the cumulative probabilities for these z-scores in the standard normal distribution table (or use a calculator):

• For \(z_{34.5} \approx 0.4545\), the cumulative probability \(P(Z < 0.4545) \approx 0.6762\) (this value can vary slightly based on the specific z-table or calculator).
• For \(z_{36.0} \approx 1.8182\), the cumulative probability \(P(Z < 1.8182) \approx 0.9656\).

4. To find the probability that the sample mean is between 34.5 and 36.0 hours:

\[ P(34.5 < \bar{X} < 36.0) = P(Z < 1.8182) - P(Z < 0.4545) \]

\[ P(34.5 < \bar{X} < 36.0) \approx 0.9656 - 0.6762 = 0.2894 \]

Thus, the probability that the sample mean life is between 34.5 and 36.0 hours is approximately 0.2894 or 28.94%.