An industrial engineer has found that the standard household light bulbs produced by a certain manufacturer have a useful life that is normally distributed with a mean of 250 hours and a variance of 2500.
(a) What is the probability that a randomly selected bulb from this production process will have a useful
life between 190 and 270 hours?
(b) Ninety percent of the bulbs have a useful life in excess of how many hours?
(c) What is the probability that the average life of 10 randomly, independently selected bulbs is less than
240 hours?
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To solve these problems, we will use the properties of the normal distribution. The normal distribution is characterized by its mean and variance.
For a normal distribution with mean μ and variance σ^2:
(a) The probability of a random variable falling within a certain range can be found by calculating the area under the curve between two points. We can use the z-score formula to find the z-scores for the values of 190 hours and 270 hours:
Z1 = (190 - μ) / σ
Z2 = (270 - μ) / σ
Substituting the given values:
Z1 = (190 - 250) / √2500 = -6 / 50 = -0.12
Z2 = (270 - 250) / √2500 = 20 / 50 = 0.4
To find the probability between these two points, we can calculate the area under the curve using a standard normal distribution table or a calculator. From the standard normal distribution table, the area corresponding to Z = 0.4 is 0.6554, and the area corresponding to Z = -0.12 is 0.4515.
Therefore, the probability that a randomly selected bulb will have a useful life between 190 and 270 hours is:
P(190 < X < 270) = P(Z1 < Z < Z2) = P(-0.12 < Z < 0.4) = 0.6554 - 0.4515 = 0.2039 or 20.39%.
(b) We need to find the value of X such that 90% of the bulbs have a longer useful life. To do this, we need to find the z-score that corresponds to the area to the left of X, which is 0.1. Using the standard normal distribution table or a calculator, we can find the z-score corresponding to an area of 0.1, which is -1.28.
To find the value of X, we can use the formula:
Z = (X - μ) / σ
Substituting the z-score and the given values:
-1.28 = (X - 250) / √2500
Simplifying the equation:
-1.28√2500 = X - 250
-3.2 = X - 250
X = 246.8
Therefore, 90% of the bulbs have a useful life in excess of 246.8 hours.
(c) The average life of 10 randomly and independently selected bulbs follows a sampling distribution, which is approximately normally distributed when the sample size is sufficiently large. The mean of the sampling distribution of the sample mean is equal to the population mean, μ, and the variance is equal to the population variance divided by the sample size, σ^2 / n.
The population variance is given as 2500, and the sample size is 10. Therefore, the variance of the sampling distribution is 2500 / 10 = 250.
We can find the z-score corresponding to a value of 240 using the formula:
Z = (X - μ) / (σ / √n)
Substituting the given values:
Z = (240 - 250) / (√250 / √10)
Z = -1 / (5 / √10)
Z = - √10 / 5
To find the probability that the average life of 10 randomly selected bulbs is less than 240 hours, we need to find the area to the left of this z-score. Using the standard normal distribution table or a calculator, we can find the area to the left of Z = - √10 / 5, which is 0.2957.
Therefore, the probability that the average life of 10 randomly and independently selected bulbs is less than 240 hours is 0.2957 or 29.57%.
To answer these questions, we need to use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetric around its mean. It is fully defined by its mean (μ) and variance (σ^2). In this case, we are given the mean (μ = 250) and variance (σ^2 = 2500).
(a) To find the probability that a randomly selected bulb will have a useful life between 190 and 270 hours, we need to calculate the area under the normal curve between these two values. We can use the z-score formula to standardize the values and then use a standard normal distribution table or a calculator to find the area.
First, we calculate the z-scores for the two values:
z1 = (190 - μ) / σ = (190 - 250) / sqrt(2500) = -6
z2 = (270 - μ) / σ = (270 - 250) / sqrt(2500) = 2
Using the standard normal distribution table or a calculator, we find the probabilities associated with these z-scores:
P(z < -6) is approximately 0, since the z-score is far in the left tail.
P(z < 2) is approximately 0.9772.
To find the probability between these two values, we subtract the probabilities:
P(190 < X < 270) = P(z < 2) - P(z < -6) = 0.9772 - 0 = 0.9772.
Therefore, the probability that a randomly selected bulb will have a useful life between 190 and 270 hours is approximately 0.9772, or 97.72%.
(b) To find the number of hours in which 90% of the bulbs have a useful life in excess of, we need to find the z-score corresponding to the 90th percentile of the distribution. This z-score is denoted as z0.9.
Using the standard normal distribution table or a calculator, we find the z-score that corresponds to a cumulative probability of 0.9:
P(z < z0.9) = 0.9
z0.9 ≈ 1.28.
We can then use the z-score formula to find the corresponding value for the useful life (X):
z0.9 = (X - μ) / σ
1.28 = (X - 250) / sqrt(2500).
Solving for X:
1.28 * sqrt(2500) = X - 250
3200 = X - 250
X ≈ 3450.
Therefore, 90% of the bulbs have a useful life in excess of approximately 3450 hours.
(c) To find the probability that the average life of 10 randomly and independently selected bulbs is less than 240 hours, we need to use the Central Limit Theorem. According to this theorem, the distribution of sample means will be approximately normally distributed regardless of the shape of the original population distribution, provided that the sample size is large enough.
The mean of the sample means (also known as the "sampling distribution") is equal to the population mean, which is given as 250. The standard deviation (σ') of the sampling distribution is given by the population standard deviation divided by the square root of the sample size.
σ' = σ / sqrt(n) = sqrt(2500) / sqrt(10)
σ' = 50 / sqrt(10)
σ' ≈ 15.8114.
To find the probability that the average life of 10 randomly selected bulbs is less than 240 hours, we need to calculate the z-score for this value:
z = (240 - μ) / σ'
z = (240 - 250) / 15.8114
z ≈ -0.6325.
Using the standard normal distribution table or a calculator, we find the cumulative probability associated with this z-score:
P(z < -0.6325) ≈ 0.2643.
Therefore, the probability that the average life of 10 randomly and independently selected bulbs is less than 240 hours is approximately 0.2643, or 26.43%.