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Answer :
Answer:
11.5%
Step-by-step explanation:
To solve this problem, we can use the knowledge that there are z score tables that calculate probabilities based on z scores. Thus, we must calculate the z score.
The z score formula is [tex]\frac{x- m}{s}[/tex] , where x is the value, m is the mean, and s is the standard deviation. We can then plug our values in to get [tex]\frac{940-1000}{50} = \frac{-60}{50} = -1.2[/tex] as our z score. Plugging this into a table where table values represent the area to the left of the z score (as we want to calculate everything under 940), we get 0.115, or 11.5% as our answer
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Final answer:
To solve this problem, we compute a z score of -1.2 for 940 hours based on a mean of 1000 hours and a standard deviation of 50 hours. Looking this score up on the standard normal distribution table, we find a probability of roughly 11.5% that a randomly selected bulb will last less than 940 hours.
Explanation:
In the realm of statistics, this is a problem calling for the calculation of a z score and using it to find a probability from the standard normal distribution. The z score is a measure of how many standard deviations an element is from the mean. Here the mean (expected life) is 1000 hours and the standard deviation is 50 hours.
First, compute the z score for 940 hours using the formula:
Z = (X - μ) / σ
where, X = 940, μ = 1000, and σ = 50. Plugging these numbers into the formula gives:
Z = (940 - 1000) / 50 = -1.2
When we look up the z score of -1.2 on the standard normal distribution table, we get the probability as 0.115. Thus, there's approximately an 11.5% chance that a randomly selected bulb would last fewer than 940 hours.
Learn more about Z Score and Probability here:
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