Final answer:
The probability of randomly selecting a cigarette with 0.669 grams of nicotine or less is calculated using the z-score method and standard normal distribution, yielding a probability of 0.1587.
Step-by-step explanation:
The question involves calculating the probability of randomly selecting a cigarette with 0.669 grams of nicotine or less when the amounts of nicotine are normally distributed with a mean of 0.967 grams and a standard deviation of 0.298 grams. To find this probability (denoted as P(X < 0.669)), we first need to find the z-score, which represents the number of standard deviations a value is from the mean. The z-score is calculated using the formula:
z = (X - μ) / σ
Where:
- X is the value of interest (0.669 grams)
- μ (mu) is the mean (0.967 grams)
- σ (sigma) is the standard deviation (0.298 grams)
Plugging in our values:
z = (0.669 - 0.967) / 0.298
z = -1.00 (rounded to two decimal places)
Now, using the standard normal distribution table, or a calculator equipped with normal distribution functions, we can find the probability associated with a z-score of -1.00. This gives us the probability we are seeking, P(X < 0.669).
The answer would typically be calculated using either a z-table or technology like a graphing calculator. For the sake of this example, let's assume the probability associated with a z-score of -1.00 is approximately 0.1587.
Therefore, the probability of randomly selecting a cigarette with 0.669 grams of nicotine or less is 0.1587 (rounded to four decimal places).