Question
The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of 1600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is taken. What is the probability that the sample mean lifetime is more than 1550 hours?
Answers
GPT-4o mini
To solve the problem of finding the probability that the sample mean lifetime of the light bulbs is more than 1550 hours, we can use the Central Limit Theorem (CLT), which states that the sampling distribution of the sample mean will be approximately normally distributed when the sample size is sufficiently large, regardless of the shape of the population distribution.
Given the following data:
- Population mean (\(\mu\)) = 1600 hours
- Population standard deviation (\(\sigma\)) = 400 hours
- Sample size (\(n\)) = 64
### Step 1: Calculate the standard error of the mean (SEM)
The standard error of the mean is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{400}{\sqrt{64}} = \frac{400}{8} = 50
\]
### Step 2: Determine the z-score for the sample mean
Next, we need to calculate the z-score for the sample mean of 1550 hours. The z-score is calculated using the formula:
\[
z = \frac{\bar{x} - \mu}{\text{SEM}}
\]
Where:
- \(\bar{x}\) = sample mean (1550 hours)
- \(\mu\) = population mean (1600 hours)
Now substituting the values:
\[
z = \frac{1550 - 1600}{50} = \frac{-50}{50} = -1
\]
### Step 3: Find the probability that the sample mean is greater than 1550
Now, we need to find the probability that the sample mean is greater than 1550 hours, which corresponds to finding \(P(Z > -1)\). Since Z is a standard normal variable, we can look up the z-score in the standard normal distribution table or use a calculator to find this probability.
The standard normal table or calculator will show that:
\[
P(Z < -1) \approx 0.1587
\]
To find \(P(Z > -1)\):
\[
P(Z > -1) = 1 - P(Z < -1) = 1 - 0.1587 = 0.8413
\]
### Conclusion
The probability that the sample mean lifetime of the light bulbs is more than 1550 hours is approximately **0.8413** or **84.13%**.
Given the following data:
- Population mean (\(\mu\)) = 1600 hours
- Population standard deviation (\(\sigma\)) = 400 hours
- Sample size (\(n\)) = 64
### Step 1: Calculate the standard error of the mean (SEM)
The standard error of the mean is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{400}{\sqrt{64}} = \frac{400}{8} = 50
\]
### Step 2: Determine the z-score for the sample mean
Next, we need to calculate the z-score for the sample mean of 1550 hours. The z-score is calculated using the formula:
\[
z = \frac{\bar{x} - \mu}{\text{SEM}}
\]
Where:
- \(\bar{x}\) = sample mean (1550 hours)
- \(\mu\) = population mean (1600 hours)
Now substituting the values:
\[
z = \frac{1550 - 1600}{50} = \frac{-50}{50} = -1
\]
### Step 3: Find the probability that the sample mean is greater than 1550
Now, we need to find the probability that the sample mean is greater than 1550 hours, which corresponds to finding \(P(Z > -1)\). Since Z is a standard normal variable, we can look up the z-score in the standard normal distribution table or use a calculator to find this probability.
The standard normal table or calculator will show that:
\[
P(Z < -1) \approx 0.1587
\]
To find \(P(Z > -1)\):
\[
P(Z > -1) = 1 - P(Z < -1) = 1 - 0.1587 = 0.8413
\]
### Conclusion
The probability that the sample mean lifetime of the light bulbs is more than 1550 hours is approximately **0.8413** or **84.13%**.