Suppose a manufacturer wants to estimate the averagelifespan of a new product they have developed. Theyrandomly select 15 products and test them until they fail.It is known that the average lifespan is 120 hours, with astandard deviation of 10 hours. Assume that the lifespanof the products follows a normal distribution. Themanufacturer wants to know the probability that theaverage lifespan of the product is greater than 130 hours.
Question
Suppose a manufacturer wants to estimate the averagelifespan of a new product they have developed. Theyrandomly select 15 products and test them until they fail.It is known that the average lifespan is 120 hours, with astandard deviation of 10 hours. Assume that the lifespanof the products follows a normal distribution. Themanufacturer wants to know the probability that theaverage lifespan of the product is greater than 130 hours.
Solution
To solve this problem, we can use the concept of the sampling distribution of the sample mean.
Step 1: Identify the parameters of the population distribution. The problem states that the average lifespan of the product is 120 hours and the standard deviation is 10 hours.
Step 2: Calculate the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of the sample size. In this case, the sample size is 15. So, the standard error of the mean is 10/sqrt(15) = 2.58 hours.
Step 3: Calculate the z-score. The z-score is the difference between the sample mean and the population mean, divided by the standard error of the mean. In this case, the sample mean is 130 hours. So, the z-score is (130-120)/2.58 = 3.88.
Step 4: Look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with this z-score. The probability that a standard normal random variable is less than 3.88 is approximately 0.9999.
Step 5: Since we want the probability that the average lifespan is greater than 130 hours, we need to subtract this probability from 1. So, the probability that the average lifespan of the product is greater than 130 hours is 1 - 0.9999 = 0.0001, or 0.01%.
Therefore, it is very unlikely that the average lifespan of the product is greater than 130 hours.
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