The lifetime of a type of electric bulb has expected value µ = 475 hours and standard deviation
σ = 60 hours.
(a) Use the central limit theorem to determine the expected value and standard deviation of the
sample mean of n such lightbulbs where n = 100, 200, or 400.
(b) You buy two such lightbulbs, use one, and when it expires, you replace it by the other. Assuming
that the lifetimes of the two lightbulbs are independent, what is the probability that the two
lightbulbs have a combined lifetime of over 1000 hours
(a) To determine the expected value and standard deviation of the sample mean, we can apply the central limit theorem. According to the central limit theorem, for a large sample size (n > 30) taken from a population with any distribution, the sample mean will be approximately normally distributed.
The expected value of the sample mean (µ¯) is equal to the population mean (µ). Thus, µ¯ = µ.
The standard deviation of the sample mean (σ¯) is equal to the population standard deviation (σ) divided by the square root of the sample size (n). Thus, σ¯ = σ / √n.
Given: µ = 475 hours, σ = 60 hours, and the sample sizes n = 100, 200, and 400.
For n = 100:
µ¯ = µ = 475 hours
σ¯ = σ / √n = 60 hours / √100 = 6 hours
For n = 200:
µ¯ = µ = 475 hours
σ¯ = σ / √n = 60 hours / √200 ≈ 4.24 hours
For n = 400:
µ¯ = µ = 475 hours
σ¯ = σ / √n = 60 hours / √400 = 3 hours
So, for n = 100, the expected value of the sample mean is 475 hours and the standard deviation is 6 hours.
For n = 200, the expected value of the sample mean is 475 hours and the standard deviation is approximately 4.24 hours.
For n = 400, the expected value of the sample mean is 475 hours and the standard deviation is 3 hours.
(b) To find the probability that the combined lifetime of the two lightbulbs is over 1000 hours, we can use the concept of the normal distribution.
If we assume that the lifetimes of the two lightbulbs are independent and normally distributed with a mean of µ and standard deviation of σ, then their combined lifetime will also be normally distributed with a mean of 2µ and a standard deviation of √(2σ^2).
Given: µ = 475 hours, σ = 60 hours.
The combined lifetime of the two lightbulbs being over 1000 hours is equivalent to the combined sample mean (2µ) being greater than 1000 hours.
Using the formula for the standard deviation of the sample mean (σ¯ = σ / √n), for n = 2 bulbs, the standard deviation becomes:
σ¯ = σ / √2 = 60 hours / √2 ≈ 42.43 hours
Now, we can standardize the desired value (1000 hours) using the formula:
z = (x - µ¯) / σ¯
Where z is the standardized value, x is the desired value, µ¯ is the mean, and σ¯ is the standard deviation.
z = (1000 - 2µ) / √(2σ^2)
= (1000 - 2 * 475) / √(2 * 60^2)
= (1000 - 950) / √(2 * 3600)
= 50 / √(7200)
≈ 1.32
Now, we can use a standard normal distribution table or software to find the probability corresponding to a z-score of 1.32. The probability that the combined lifetime of the two lightbulbs is over 1000 hours is the area under the curve to the right of the z-score.
Using a standard normal distribution table or software, we find that the probability is approximately 0.0934, or 9.34%.
To solve this problem, we will use the central limit theorem and the properties of expected value and standard deviation.
(a) Expected Value and Standard Deviation of Sample Mean:
According to the central limit theorem, the distribution of sample means approaches a normal distribution as the sample size increases. We can calculate the expected value and standard deviation of the sample mean using the formulas:
Expected Value (µ') = µ (same as the population mean)
Standard Deviation (σ') = σ / √n (where n is the sample size)
For n = 100:
µ' = µ = 475 hours
σ' = σ / √n = 60 hours / √100 = 60 hours / 10 = 6 hours
For n = 200:
µ' = µ = 475 hours
σ' = σ / √n = 60 hours / √200 = 60 hours / 14.14 ≈ 4.24 hours
For n = 400:
µ' = µ = 475 hours
σ' = σ / √n = 60 hours / √400 = 60 hours / 20 = 3 hours
(b) Probability of Combined Lifetime:
Assuming the lifetimes of the two light bulbs are independent, we can add their lifetimes to calculate the combined lifetime.
Let's assume X and Y are the lifetimes (in hours) of the two light bulbs. We want to find P(X + Y > 1000).
Since X and Y are independent, the sum of their lifetimes will follow a normal distribution with mean equal to the sum of the individual means (2µ) and standard deviation equal to the square root of the sum of the individual variances (sqrt(σ^2 + σ^2)).
Mean (µ_sum) = 2 * µ = 2 * 475 = 950 hours
Standard Deviation (σ_sum) = sqrt(σ^2 + σ^2) = sqrt(60^2 + 60^2) = sqrt(7200 + 7200) ≈ sqrt(14400) ≈ 120 hours
To calculate the probability:
P(X + Y > 1000) = 1 - P(X + Y ≤ 1000)
Now we need to standardize the value of 1000 using the distribution parameters:
Z = (1000 - µ_sum) / σ_sum
Then, we can find the probability using the standard normal distribution.
Finally, calculate P(X + Y > 1000) using the standard normal distribution table or software.
(a) Sample mean is expected to be the same. (Standard Error of the mean) SEm = SD/√n
(b) Assume equal times. One bulb = 500 hours.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.