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Answer :
Final answer:
Given the screen's normal distribution with a mean of 6 and variance of 0.01, the probability that the screen length falls between 5.95 and 6.2 inches—calculated using z-scores and the normal distribution table—is about 66.87%.
Explanation:
In this scenario, we're dealing with a normal distribution —specifically, the length of tablet screens produced by a manufacturer. We know that the mean screen length is 6 inches, and the variance of the manufacturing process is 0.01. This gives a standard deviation of the square root of the variance, which is 0.1 inches.
To calculate the probability that a screen length falls between 5.95 and 6.2 inches, we need to standardize these lengths—turn them into z-scores—using the mean and standard deviation. The z-score formula is (X - μ) / σ, where X is the measurement we're interested in, μ is the mean, and σ is the standard deviation. For 5.95, that gives us (5.95 - 6) / 0.1 = -0.5, and for 6.2, we get (6.2 - 6) / 0.1 = 2.
Now, we look up these z-scores on a standard normal distribution table. The value corresponding to 2 is approximately 0.9772—this means about 97.72% of screen lengths are less than 6.2 inches. The value corresponding to -0.5 is approximately 0.3085, or about 30.85% of screen lengths are less than 5.95 inches.
To find the probability that a screen length falls within our customer's acceptable range, we subtract the lower z-score from the higher one: 0.9772 - 0.3085 = 0.6687, or approximately 66.87%.
So, there's a 66.87% probability that a randomly selected screen length will meet the customer's requirements.
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