the life of light bulbs is distributed normally. The variance of the life time is 225 and the mean lifetime of a bulb is 590 hours. Find the probability of a bulb lasting for at most 603 hours. Yeah, WTF??? Why me!??!
The diameters of ball bearings are distributed normally. The mean diameter is
87
millimeters and the standard deviation is
6
millimeters. Find the probability that the diameter of a selected bearing is greater than
84
millimeters. Round your answer to four decimal places.
To find the probability of a bulb lasting for at most 603 hours, we can use the concept of standard deviation and z-scores in a normal distribution.
Step 1: Calculate the standard deviation (σ) from the variance (var).
The variance (var) is given as 225. The standard deviation (σ) is the square root of the variance: σ = √var = √225 = 15.
Step 2: Calculate the z-score.
The z-score measures the number of standard deviations a value is from the mean. We can calculate the z-score using the formula: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability of a bulb lasting for at most 603 hours, so X = 603 hours. The mean lifetime of a bulb is given as μ = 590 hours, and the standard deviation is σ = 15 hours. Plugging these values into the formula, we get:
z = (603 - 590) / 15 ≈ 0.87
Step 3: Find the probability.
Now, we need to find the probability of a bulb lasting for at most 603 hours, which is equivalent to finding the probability of having a z-score of 0.87 or less. We can use a standard normal distribution table or a calculator to find this probability.
Using the standard normal distribution table, we look for the closest corresponding value to 0.87, which is 0.8078. This means that the probability of a bulb lasting for at most 603 hours is approximately 0.8078, or 80.78%.
Remember, the process outlined above can be used to solve similar problems involving normal distributions and finding probabilities using z-scores.
marshall
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 from the Z score.
Why not you?