SOLUTION: "An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 802 hours and a standard deviation of 93 hour

Algebra ->  Probability-and-statistics -> SOLUTION: "An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 802 hours and a standard deviation of 93 hour      Log On


   

Question 1191348: "An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 802 hours and a standard deviation of 93 hours. Find the probability that a random samples of 23 bulbs will have an average life between 793 and 817 hours."
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to solve this problem:
1. **Find the standard error of the mean:**
The standard error of the mean (SEM) is the standard deviation of the sample means. It's calculated as:
SEM = σ / √n
Where:
* σ is the population standard deviation (93 hours)
* n is the sample size (23 bulbs)
SEM = 93 / √23 ≈ 19.38
2. **Calculate the z-scores:**
We need to convert the given average life values (793 and 817 hours) into z-scores. The z-score tells us how many standard errors a particular sample mean is away from the population mean.
z = (x - μ) / SEM
Where:
* x is the sample mean
* μ is the population mean (802 hours)
* For x = 793 hours:
z₁ = (793 - 802) / 19.38 ≈ -0.47
* For x = 817 hours:
z₂ = (817 - 802) / 19.38 ≈ 0.78
3. **Find the probabilities:**
Use a z-table or calculator to find the area under the normal curve between these two z-scores. This represents the probability that the sample mean will fall between 793 and 817 hours.
* Find the probability associated with z₂ = 0.78: P(z < 0.78) ≈ 0.7823
* Find the probability associated with z₁ = -0.47: P(z < -0.47) ≈ 0.3192
4. **Calculate the probability between the two values:**
Subtract the smaller probability from the larger probability:
P(-0.47 < z < 0.78) = P(z < 0.78) - P(z < -0.47)
P(-0.47 < z < 0.78) = 0.7823 - 0.3192 ≈ 0.4631
Therefore, the probability that a random sample of 23 bulbs will have an average life between 793 and 817 hours is approximately 0.4631 or 46.31%.