Cell Phone Lifetimes A recent study of the lifetimes of cell phones found the average is 23.2 months. The standard deviation is 3.1 months. If a company provides its 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 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. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places. P(X<23.8)=

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**Cell Phone Lifetimes**

A recent study of the lifetimes of cell phones found the average is 23.2 months. The standard deviation is 3.1 months. If a company provides 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 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. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places.

\[ 
P\left(\overline{X} < 23.8\right) = \square
\]

### Instructions:

To solve this problem, follow these steps:

1. **Calculate the Standard Error (SE):**
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]
   - Standard deviation (\( \sigma \)) = 3.1 months
   - Sample size (\( n \)) = 33 employees

2. **Determine the Z-score:**
\[ Z = \frac{\overline{X} - \mu}{SE} \]
   - Mean (\( \mu \)) = 23.2 months
   - Target mean (\( \overline{X} \)) = 23.8 months

3. **Find the Probability:**
   - Once the Z-score is calculated, use a Z-table or a calculator to find the corresponding probability.

### Example Calculation:

1. **Standard Error (SE):**
\[ 
\text{SE} = \frac{3.1}{\sqrt{33}} \approx 0.5394 
\]

2. **Z-score:**
\[ 
Z = \frac{23.8 - 23.2}{0.5394} \approx 1.113 
\]

3. **Probability:**
   - Using a Z-table or calculator, find the probability corresponding to Z = 1.113. 

### TI-83 Plus/TI-84 Plus Calculator Steps:

1. **Press 2nd/DISTR**
2. **Select normalcdf**
3. **Enter (-E99, 23.8, 23.2, 0.5394)** and press ENTER

This will give you the probability that the average lifetime of 33 cell phones is less than 23.8 months
Transcribed Image Text:**Cell Phone Lifetimes** A recent study of the lifetimes of cell phones found the average is 23.2 months. The standard deviation is 3.1 months. If a company provides 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 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. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places. \[ P\left(\overline{X} < 23.8\right) = \square \] ### Instructions: To solve this problem, follow these steps: 1. **Calculate the Standard Error (SE):** \[ \text{SE} = \frac{\sigma}{\sqrt{n}} \] - Standard deviation (\( \sigma \)) = 3.1 months - Sample size (\( n \)) = 33 employees 2. **Determine the Z-score:** \[ Z = \frac{\overline{X} - \mu}{SE} \] - Mean (\( \mu \)) = 23.2 months - Target mean (\( \overline{X} \)) = 23.8 months 3. **Find the Probability:** - Once the Z-score is calculated, use a Z-table or a calculator to find the corresponding probability. ### Example Calculation: 1. **Standard Error (SE):** \[ \text{SE} = \frac{3.1}{\sqrt{33}} \approx 0.5394 \] 2. **Z-score:** \[ Z = \frac{23.8 - 23.2}{0.5394} \approx 1.113 \] 3. **Probability:** - Using a Z-table or calculator, find the probability corresponding to Z = 1.113. ### TI-83 Plus/TI-84 Plus Calculator Steps: 1. **Press 2nd/DISTR** 2. **Select normalcdf** 3. **Enter (-E99, 23.8, 23.2, 0.5394)** and press ENTER This will give you the probability that the average lifetime of 33 cell phones is less than 23.8 months
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