Final answer:
The probability of randomly selecting a cigarette with 0.518g of nicotine or less is determined by calculating the z-score, which in this case does not match exactly with the provided options. The closest option with the calculated z-score is (a) P(X<0.518g)=0.2385.
Step-by-step explanation:
To find the probability of randomly selecting a cigarette with 0.518g of nicotine or less from a normally distributed amount of nicotine with a mean of 0.916g and a standard deviation of 0.306g, we need to normalize the value and find the corresponding area under the normal curve. This is done by calculating the z-score.
The z-score is calculated as follows:
Z = (X - mean) / standard deviation
Z = (0.518 - 0.916) / 0.306 = -1.3 approximately
Using a z-table or a calculator with normal distribution functions, we find the probability that Z is less than -1.3. This gives us a probability of approximately 0.0968.
However, none of the options provided (a) 0.2385, (b) 0.3551, (c) 0.6826, and (d) 0.8413 exactly match the calculated probability. Therefore, we should check for potential errors or consider the possibility that the correct answer may not be listed, provided that my computations are indeed correct. Given the provided context and information, an exact match to the options is not available. For instructional purposes, the closest answer to our calculation is (a) P(X<0.518g)=0.2385.