Battery manufacturers compete on the basis of the amount of time their products last in cameras and toys. A manufacturer of alkaline batteries has observed that it's batteries last for an average of 26 hours when used in a toy racing car. The amount of time is normally distributed with a standard deviation of 2.5 hours. What is the probability that the printer produces more than 12,000 pages before this cartridge must be replaced? What is the probability that the printer produces fewer than 10,000 pages?
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check your average and standard deviation.
To determine the probability that the printer produces more than 12,000 pages, we need to find the z-score and use the standard normal distribution table.
Step 1: Calculate the z-score
The z-score formula is (X - μ) / σ, where X is the desired value, μ is the mean, and σ is the standard deviation.
In this case, X = 12,000, μ = 26 hours, and σ = 2.5 hours.
z-score = (12,000 - 26) / 2.5 = 4,948
Step 2: Lookup the z-score in the standard normal distribution table
Since the standard normal distribution table provides probabilities for z-scores less than the given value, we need to find the complementary probability for z = 4,948.
The standard normal distribution table may not list an exact value of 4,948, but we can approximate it by using the value of 4.9.
The standard normal distribution table provides the probability for z ≤ 4.9, which is 1.0. To find the probability for z > 4.9, subtract 1.0 from 1.0.
Therefore, the probability that the printer produces more than 12,000 pages is approximately 0.
To determine the probability that the printer produces fewer than 10,000 pages, we repeat the same steps.
Step 1: Calculate the z-score
Using the same formula, we have X = 10,000, μ = 26 hours, and σ = 2.5 hours.
z-score = (10,000 - 26) / 2.5 = 3,990
Step 2: Lookup the z-score in the standard normal distribution table
Since the standard normal distribution table provides probabilities for z-scores less than the given value, we can use the value of 3.9 for approximation.
The standard normal distribution table provides the probability for z ≤ 3.9, which is approximately 0.9999.
Therefore, the probability that the printer produces fewer than 10,000 pages is approximately 0.9999.
Please note that these probability values are approximations given the use of the standard normal distribution table.