The following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones:Suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. How likely is it that a randomly chosen battery lasts longer than 10.4 hours? 0.50 0.16 0.05 0.025
Question
The following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones:Suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. How likely is it that a randomly chosen battery lasts longer than 10.4 hours? 0.50 0.16 0.05 0.025
Solution
To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.
Step 1: Identify the mean (μ) and the standard deviation (σ). From the problem, we know that μ = 8 and σ = 1.2.
Step 2: Identify the value for which we want to find the probability (X). In this case, X = 10.4.
Step 3: Calculate the Z-score. The formula for the Z-score is Z = (X - μ) / σ. Substituting the given values, we get Z = (10.4 - 8) / 1.2 = 2.
Step 4: Look up the Z-score in the Z-table to find the probability. The Z-table tells us the probability that a randomly chosen battery lasts less than 10.4 hours. But we want to find the probability that a battery lasts longer than 10.4 hours.
The Z-table value for Z=2 is 0.9772, which means the probability that a battery lasts less than 10.4 hours is 0.9772 or 97.72%.
Step 5: Since we want the probability that a battery lasts longer than 10.4 hours, we subtract the value found from 1 (because the total probability is 1). So, 1 - 0.9772 = 0.0228 or 2.28%.
So, the likelihood that a randomly chosen battery lasts longer than 10.4 hours is 2.28%. The closest answer among the options given is 0.025 or 2.5%.
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