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A recent study of the lifetime of cell phones found the average is 24.3 months. The standard deviation is 2.6 months. If a company provides its 34 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.0 months. Assume cell phone life is a normally distributed variable, the sample is taken from a large population, and the correction factor can be ignored. Round the final answer to 4 decimal places and intermediate z-value calculations to 2 decimal places.



$P(\bar{X}<23.0) =$

Verified Answer :

  • Calculate the standard error of the mean: \(\sigma_{\bar{x}} = \frac{2.6}{\sqrt{34}} \approx 0.4459\).
  • Find the \(z\)-score: \(z = \frac{23.0 - 24.3}{0.4459} \approx -2.92\).
  • Determine the probability corresponding to the \(z\)-score using the normal distribution: \(P(Z < -2.92) = 0.0018\).
  • The final probability is \(\boxed{0.0018}\).

Explanation

  1. Understand the problem and parameters

We are analyzing the lifetimes of cell phones, which are assumed to follow a normal distribution. We are given the following parameters from the study:

  • Population mean (\(\mu\)): \(24.3\) months
  • Population standard deviation (\(\sigma\)): \(2.6\) months
  • Sample size (\(n\)): \(34\)

We want to find the probability that the sample mean (\(\bar{X}\)) for a group of \(34\) phones is less than \(23.0\) months, denoted as \(P(\bar{X} < 23.0)\). Because we are dealing with a sample mean, we use the Central Limit Theorem to determine the sampling distribution of \(\bar{X}\), which will be normal with the same mean \(\mu\) and a standard error \(\sigma_{\bar{x}}\).

  1. Calculate the standard error of the mean

To work with the sampling distribution, we first need to calculate the standard error of the mean (\(\sigma_{\bar{x}}\)). This is calculated as the population standard deviation divided by the square root of the sample size:
\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.6}{\sqrt{34}} \approx \frac{2.6}{5.83095} \approx 0.445896\]
Rounding to a reasonable level, we use \(\sigma_{\bar{x}} \approx 0.4459\).

  1. Calculate the z-score

Next, we calculate the \(z\)-score for our sample mean of \(\bar{X} = 23.0\). The \(z\)-score indicates how many standard errors the sample mean is away from the population mean:
\[z = \frac{\bar{X} - \mu}{\sigma_{\bar{x}}} = \frac{23.0 - 24.3}{0.4459} \approx \frac{-1.3}{0.4459} \approx -2.91547\]
Rounding to two decimal places as requested, we get \(z = -2.92\).

  1. Determine the probability

Now, we find the probability \(P(Z < -2.92)\) using the standard normal distribution. This represents the area under the curve to the left of \(z = -2.92\). Using standard normal distribution tables or calculation, we find:
\[P(Z < -2.92) \approx 0.0018\]
This means there is a very small probability, approximately \(0.18\%\), that the sample mean lifetime of the \(34\) phones will be less than \(23.0\) months.

Examples

In quality control, this method is used to determine if a batch of products meets performance standards. For example, a manufacturer can assess the probability that a random sample of produced lightbulbs has a mean lifespan lower than expected, helping to decide if the production line needs calibration. By using the Central Limit Theorem, you can estimate these risks without testing every single item in the entire production population.

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