A tobacco company claims that the nicotine content of its "light" cigarettes has a mean of 1.58 milligrams and a standard deviation of 0.59 milligrams. What is the probability that 50 randomly selected light cigarettes from this company will have a total combined nicotine content of 83 milligrams or less, assuming the company's claims to be true?Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.
Question
A tobacco company claims that the nicotine content of its "light" cigarettes has a mean of 1.58 milligrams and a standard deviation of 0.59 milligrams. What is the probability that 50 randomly selected light cigarettes from this company will have a total combined nicotine content of 83 milligrams or less, assuming the company's claims to be true?Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.
Solution
To solve this problem, we will use the Central Limit Theorem, which states that the sum or average of a large number of measurements will follow a normal distribution, regardless of the shape of the original measurement distribution, as long as the measurements are independent.
Step 1: Identify the parameters The mean (μ) = 1.58 milligrams The standard deviation (σ) = 0.59 milligrams The sample size (n) = 50 cigarettes The total combined nicotine content we are interested in (X) = 83 milligrams
Step 2: Calculate the mean and standard deviation for the sum of 50 cigarettes The mean for the sum (μ_sum) = n * μ = 50 * 1.58 = 79 milligrams The standard deviation for the sum (σ_sum) = sqrt(n) * σ = sqrt(50) * 0.59 = 4.1669 milligrams
Step 3: Standardize the total combined nicotine content to a z-score The z-score is calculated as (X - μ_sum) / σ_sum = (83 - 79) / 4.1669 = 0.9594
Step 4: Look up the z-score in a standard normal distribution table or use a calculator to find the probability The probability that a standard normal variable is less than 0.9594 is approximately 0.8315.
So, the probability that 50 randomly selected light cigarettes from this company will have a total combined nicotine content of 83 milligrams or less is approximately 0.831 or 83.1%, assuming the company's claims to be true.
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