The average age of a vehicles registered in Kenya is 8 years or 96 months, assume the standard deviation is 16 months. If a random sample of 36 vehicle is selected, find the probability that the selected age is between 90 and 100 months
Question
The average age of a vehicles registered in Kenya is 8 years or 96 months, assume the standard deviation is 16 months. If a random sample of 36 vehicle is selected, find the probability that the selected age is between 90 and 100 months
Solution
To solve this problem, we will use the concept of Z-scores in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.
Step 1: Identify the given values. The mean (μ) = 96 months The standard deviation (σ) = 16 months Sample size (n) = 36 vehicles We are looking for the probability that the age of the vehicle is between 90 and 100 months.
Step 2: Calculate the Z-scores for 90 and 100 months. The formula for the Z-score is Z = (X - μ) / (σ/√n)
For X = 90 months, the Z-score is Z = (90 - 96) / (16/√36) = -6 / (16/6) = -2.25
For X = 100 months, the Z-score is Z = (100 - 96) / (16/√36) = 4 / (16/6) = 1.5
Step 3: Find the probability that the Z-score is between -2.25 and 1.5. We can look up these Z-scores in the Z-table (also known as the standard normal distribution table) to find the corresponding probabilities.
The probability for Z = -2.25 is 0.0122 (this value represents the probability that the age is less than 90 months).
The probability for Z = 1.5 is 0.9332 (this value represents the probability that the age is less than 100 months).
Step 4: Subtract the two probabilities to find the probability that the age is between 90 and 100 months. P(90 < X < 100) = P(X < 100) - P(X < 90) = 0.9332 - 0.0122 = 0.921
So, the probability that the age of a randomly selected vehicle is between 90 and 100 months is approximately 0.921 or 92.1%.
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